- Look for one power whose square produces the highest power. - Identify the expression that appears both to the first and second power.

1. The powers are \(x^4=(x^2)^2\) and \(x^2\). 2. Replacing the repeated quantity \(x^2\) with a new variable makes the equation quadratic.

Let \(u=x^2\).

- Compare the exponents \(6\) and \(3\). - Look for a quantity whose square gives the highest-power term.

1. The highest power satisfies \(x^6=(x^3)^2\). 2. The repeated quantity is \(x^3\), so replacing it with a new variable makes the equation quadratic.

Let \(u=x^3\).

- Look for an entire expression that repeats, not just a power of \(x\). - One occurrence of that expression is squared.

1. The expression \(x^2-1\) appears once squared and once to the first power. 2. Replacing that entire repeated expression with a new variable produces a quadratic equation.

Let \(u=x^2-1\).

- Express \(x\) as a square of the other expression that appears. - Remember the output restriction for a principal square root.

1. Since \(x=(\sqrt{x})^2\), the equation contains the same quantity to the second and first powers. 2. Let \(u=\sqrt{x}\). 3. Because \(u\) is a principal square root, \(u\ge0\).

Let \(u=\sqrt{x}\), with \(u\ge0\).

- How can the highest power be expressed using a square of the repeated lower power? - Replace every occurrence of the repeated expression, not only one of them.

1. Since \(u=x^2\), then \(x^4=u^2\). 2. Replacing these powers gives \(u^2-5u+6=0\).

\(u^2-5u+6=0\)

- Put the original expression for \(u\) back into the equation. - Which real numbers have the given square?

1. Back-substitute to obtain \(x^2=9\). 2. Both \(x=3\) and \(x=-3\) square to \(9\).

\(x=-3\) or \(x=3\).

- What do you know about the sign of a real number squared? - Is every solution for the substituted letter necessarily allowed for the original expression?

1. Back-substitution gives \(x^2=-4\). 2. A real square is never negative, so this branch gives no real solution.

No. There is no real \(x\) with \(x^2=-4\).

- Use the rule above the table to calculate each missing output. - What happens to a negative sign when you square a number?

1. At \(x=-3\), \(u=(-3)^2=9\). 2. At \(x=3\), \(u=3^2=9\). 3. The values agree because opposite numbers have the same square.

Both blanks are \(9\). The inputs \(-3\) and \(3\) have the same square.

- Replace \(x^2\) with a new variable. - Solve the resulting quadratic equation. - Substitute back to the original variable. - Remember that each positive value of \(x^2\) gives two real values of \(x\).

1. Let \(u = x^2\). Then the equation becomes \(u^2 - 13u + 36 = 0\). 2. Factor: \((u - 4)(u - 9) = 0\), so \(u = 4\) or \(u = 9\). 3. Substitute back. From \(x^2 = 4\), \(x = \pm 2\). From \(x^2 = 9\), \(x = \pm 3\). 4. The real solution set is \(\{-3, -2, 2, 3\}\).

\(\{-3, -2, 2, 3\}\)

- Replace \(x^2\) with a new variable. - Solve the resulting quadratic equation carefully. - Check which values of the substituted variable can equal a real square. - Distinguish solutions for the new variable from solutions for \(x\).

1. Let \(u = x^2\). Then the equation becomes \(u^2 + 5u - 36 = 0\). 2. Factor: \((u + 9)(u - 4) = 0\), so \(u = -9\) or \(u = 4\). 3. Substitute back. The equation \(x^2 = -9\) has no real solutions. From \(x^2 = 4\), \(x = \pm 2\). 4. The real solution set is \(\{-2, 2\}\).

\(\{-2, 2\}\)

- Replace \(x^2\) with a new variable. - Solve the resulting quadratic equation. - Substitute back to \(x\). - Include both signs when taking square roots.

1. Set \(4x^4 - 17x^2 + 4 = 0\). 2. Let \(u = x^2\). Then \(4u^2 - 17u + 4 = 0\). 3. Factor: \((4u - 1)(u - 4) = 0\), so \(u = \frac{1}{4}\) or \(u = 4\). 4. Substitute back. From \(x^2 = \frac{1}{4}\), \(x = \pm\frac{1}{2}\). From \(x^2 = 4\), \(x = \pm 2\).

\(x = -2, -\frac{1}{2}, \frac{1}{2}, 2\)

- Replace \(x^2\) with a new variable. - Solve the resulting quadratic equation. - Substitute back to the original variable. - Include both square roots of each positive value.

1. Let \(u = x^2\). Then \(u^2 - 26u + 25 = 0\). 2. Factor: \((u - 1)(u - 25) = 0\), so \(u = 1\) or \(u = 25\). 3. Substitute back. From \(x^2 = 1\), \(x = \pm 1\). From \(x^2 = 25\), \(x = \pm 5\). 4. The real solution set is \(\{-5, -1, 1, 5\}\).

\(\{-5, -1, 1, 5\}\)

- Simplify the quadratic equation after substituting. - A real square cannot be negative. - Check whether every value of the substituted variable gives a real solution. - Leave irrational solutions in exact radical form.

1. Let \(u = x^2\). Then \(2u^2 + 6u - 80 = 0\). 2. Divide by \(2\): \(u^2 + 3u - 40 = 0\). 3. Factor: \((u - 5)(u + 8) = 0\), so \(u = 5\) or \(u = -8\). 4. The equation \(x^2 = 5\) gives \(x = \pm\sqrt{5}\). The equation \(x^2 = -8\) has no real solutions.

\(\{-\sqrt{5}, \sqrt{5}\}\)

- Replace \(x^2\) with a new variable. - Solve the resulting quadratic equation. - Use the zeros to write linear factors. - Apply the difference-of-squares pattern when helpful.

1. Let \(u = x^2\). Then solve \(u^2 - 29u + 100 = 0\). 2. Factor: \((u - 4)(u - 25) = 0\), so \(u = 4\) or \(u = 25\). 3. Substitute back. From \(x^2 = 4\), \(x = \pm 2\). From \(x^2 = 25\), \(x = \pm 5\). 4. Therefore, \(A(x) = (x - 5)(x + 5)(x - 2)(x + 2)\).

\(A(x) = (x - 5)(x + 5)(x - 2)(x + 2)\)

- Replace \(x^2\) with a new variable. - Substitute back after solving the quadratic equation. - A real square cannot be negative. - Include every real value in each solution set.

1. For a), let \(u = x^2\). Then \(u^2 - 11u + 18 = 0\), so \(u = 2\) or \(u = 9\). Therefore, \(x = \pm\sqrt{2}\) or \(x = \pm 3\). 2. For b), let \(u = x^2\). Then \(u^2 + 6u + 5 = 0\), so \(u = -1\) or \(u = -5\). Neither value can equal \(x^2\) for real \(x\), so there are no real solutions. 3. For c), let \(u = x^2\). Then \(2u^2 - 7u - 4 = 0\), so \(u = 4\) or \(u = -\frac{1}{2}\). Only \(u = 4\) gives real values, so \(x = \pm 2\).

a) \(\{-3, -\sqrt{2}, \sqrt{2}, 3\}\) b) \(\varnothing\) c) \(\{-2, 2\}\)

- Use a substitution to rewrite the fourth-degree equation as a quadratic equation. - After solving the quadratic equation, substitute back for the original variable. - A fourth-degree polynomial can have at most four real zeros.

1. Set \(f(x) = 0\): \(x^4 - 20x^2 + 64 = 0\). 2. Let \(u = x^2\). Then \(u^2 - 20u + 64 = 0\). 3. Factor: \((u - 16)(u - 4) = 0\), so \(u = 16\) or \(u = 4\). 4. Back-substitute. From \(x^2 = 16\), \(x = \pm 4\). From \(x^2 = 4\), \(x = \pm 2\).

The zeros are \(x = -4\), \(x = -2\), \(x = 2\), and \(x = 4\).

- Identify the repeated expression whose square gives the first term. - Solve the resulting quadratic before returning to \(x\). - Undo a cube-root substitution by using the corresponding power.

1. Let \(u=x^{\frac{1}{3}}\). Then \(x^{\frac{2}{3}}=u^2\). 2. The equation becomes \(u^2-5u+6=0\). 3. Factor: \((u-2)(u-3)=0\), so \(u=2\) or \(u=3\). 4. Since \(u=\sqrt[3]{x}\), the corresponding values are \(x=2^3=8\) and \(x=3^3=27\).

\(x=8\) or \(x=27\)

- Which power of \(x\) is unchanged when you replace \(x\) by its opposite? - Substitute the common squared value to make an equation for the missing coefficient.

1. With \(u=x^2\), both inputs have \(u=4\). 2. The substituted equation \(u^2+cu-8=0\) must be true at \(u=4\), so \(16+4c-8=0\), giving \(c=-2\). 3. Both \(2\) and \(-2\) have square \(4\) and satisfy the equation.

\(c=-2\). Both inputs give \(u=x^2=4\), and \(u^2-2u-8=0\) holds at \(u=4\).

- Which repeated power is squared to make the highest power in each equation? - After choosing a substitution, can every variable term be written using only \(u\) and \(u^2\)?

1. In a), \(x^4=(x^2)^2\), so using \(u=x^2\) gives \(2u^2+3u-5=0\). 2. In b), \(x^6=(x^3)^2\), so using \(u=x^3\) gives \(u^2-4u+3=0\). 3. Each substitution works because the higher variable power is the square of the lower one.

a) \(u=x^2\), giving \(2u^2+3u-5=0\). b) \(u=x^3\), giving \(u^2-4u+3=0\).

- Use a substitution that turns the equation into a quadratic. - Substitute back after solving the quadratic equation. - Keep the original leading coefficient in the fully factored form.

1. Set the function equal to zero: \(2x^4 - 10x^2 + 8 = 0\). 2. Divide by \(2\): \(x^4 - 5x^2 + 4 = 0\). 3. Let \(u = x^2\). Then \(u^2 - 5u + 4 = 0\). 4. Factor: \((u - 1)(u - 4) = 0\), so \(u = 1\) or \(u = 4\). 5. Substitute back. From \(x^2 = 1\), \(x = \pm 1\). From \(x^2 = 4\), \(x = \pm 2\). 6. Preserve the leading coefficient when writing the factorization: \(g(x) = 2(x - 1)(x + 1)(x - 2)(x + 2)\).

Zeros: \(x = -2, -1, 1, 2\) Fully factored form: \(g(x) = 2(x - 1)(x + 1)(x - 2)(x + 2)\)

- Simplify the equation by dividing by a common factor. - Replace \(x^2\) with a new variable. - Check whether each substituted value can equal a real square.

1. Set \(3x^4 - 15x^2 - 108 = 0\). 2. Divide by \(3\): \(x^4 - 5x^2 - 36 = 0\). 3. Let \(u = x^2\). Then \(u^2 - 5u - 36 = 0\). 4. Factor: \((u - 9)(u + 4) = 0\), so \(u = 9\) or \(u = -4\). 5. The equation \(x^2 = 9\) gives \(x = \pm 3\). The equation \(x^2 = -4\) has no real solutions. 6. Therefore, the x-intercepts are \((-3, 0)\) and \((3, 0)\).

\((-3, 0)\) and \((3, 0)\)

- Apply the zero-product property when the function is already factored. - Factor out a common power of \(x\) when possible. - Look for an equation in quadratic form after factoring. - A fifth-degree polynomial can have at most five real zeros.

1. For \(h\), use the zero-product property. From \(x^2 - 2 = 0\), \(x = \pm\sqrt{2}\). From \(x^2 - 25 = 0\), \(x = \pm 5\). 2. For \(k\), factor out \(x\): \(k(x) = x(x^4 - 13x^2 + 36)\). Thus, \(x = 0\) is one zero. 3. To solve \(x^4 - 13x^2 + 36 = 0\), let \(u = x^2\). Then \(u^2 - 13u + 36 = 0\). 4. Factor: \((u - 4)(u - 9) = 0\), so \(u = 4\) or \(u = 9\). 5. Substitute back to get \(x = \pm 2\) and \(x = \pm 3\).

a) \(x = -5, -\sqrt{2}, \sqrt{2}, 5\) b) \(x = -3, -2, 0, 2, 3\)

- Substitute \(x = 3\) directly into the function. - Express \(x^6\) in terms of \(u = x^3\). - Remember that an odd root of a negative number is real. - Solve each equation after substituting back.

1. Evaluate the function: \(g(3) = 3^6 - 26(3^3) - 27 = 729 - 702 - 27 = 0\). Therefore, \(x = 3\) is a zero. 2. Let \(u = x^3\). Then \(x^6 = u^2\), so the equation becomes \(u^2 - 26u - 27 = 0\). 3. Factor: \((u - 27)(u + 1) = 0\), so \(u = 27\) or \(u = -1\). 4. Substitute back. From \(x^3 = 27\), \(x = 3\). From \(x^3 = -1\), \(x = -1\). 5. The only additional real zero is \(x = -1\).

\(g(3) = 0\), and the other real zero is \(x = -1\).

- Set the two function rules equal. - Rewrite the resulting equation in quadratic form. - Substitute for \(x^2\), then substitute back. - Use the constant function to determine every y-coordinate.

1. Set the function values equal: \(x^4 - 10x^2 = -9\). 2. Rewrite as \(x^4 - 10x^2 + 9 = 0\). 3. Let \(u = x^2\). Then \(u^2 - 10u + 9 = 0\). 4. Factor: \((u - 1)(u - 9) = 0\), so \(u = 1\) or \(u = 9\). 5. Substitute back. From \(x^2 = 1\), \(x = \pm 1\). From \(x^2 = 9\), \(x = \pm 3\). 6. Every intersection has y-coordinate \(-9\).

\((-3, -9)\), \((-1, -9)\), \((1, -9)\), and \((3, -9)\)

- Factor out the common numerical factor first. - Replace \(x^2\) with a new variable. - Use the zeros to write linear factors. - Preserve the original leading coefficient.

1. Factor out \(2\): \(f(x) = 2(x^4 - 13x^2 + 36)\). 2. Let \(u = x^2\). Then solve \(u^2 - 13u + 36 = 0\). 3. Factor: \((u - 4)(u - 9) = 0\), so \(u = 4\) or \(u = 9\). 4. Substitute back. The real zeros are \(x = \pm 2\) and \(x = \pm 3\). 5. Include the original leading coefficient: \(f(x) = 2(x - 2)(x + 2)(x - 3)(x + 3)\).

\(f(x) = 2(x - 2)(x + 2)(x - 3)(x + 3)\)

- Rewrite the quartic as an equation in \(x^2\). - Substitute back after solving the quadratic equation. - Use each zero to form a linear factor. - Compare the number of distinct zeros with Zara's claim.

1. Let \(u=x^2\). Then \(u^2-10u+9=0\). 2. Factor: \((u-1)(u-9)=0\), so \(u=1\) or \(u=9\). 3. Substitute back. From \(x^2=1\), \(x=\pm1\). From \(x^2=9\), \(x=\pm3\). 4. There are four distinct real zeros, so the claim is true. 5. The complete factorization is \(p(x)=(x-1)(x+1)(x-3)(x+3)\).

a) The claim is true. The zeros are \(x=-3,-1,1,3\). b) \(p(x)=(x-1)(x+1)(x-3)(x+3)\)

- Not every solution for the substituted variable produces a real value of \(x\). - A real square cannot be negative. - Keep the leading coefficient when solving the first quadratic equation. - Check the final values in the original equations.

1. For equation 1, let \(u = x^2\). Then \(4u^2 - 37u + 9 = 0\). 2. Factor: \((4u - 1)(u - 9) = 0\), so \(u = \frac{1}{4}\) or \(u = 9\). 3. Substitute back to get \(x = \pm\frac{1}{2}\) and \(x = \pm 3\). There are four real solutions. 4. For equation 2, let \(u = x^2\). Then \(u^2 - 3u - 4 = 0\). 5. Factor: \((u - 4)(u + 1) = 0\), so \(u = 4\) or \(u = -1\). 6. Only \(u = 4\) gives real values of \(x\), namely \(x = \pm 2\). There are two real solutions.

1) \(\{-3, -\frac{1}{2}, \frac{1}{2}, 3\}\); 4 real solutions 2) \(\{-2, 2\}\); 2 real solutions

- Factor out the greatest common factor first. - Replace \(x^2\) with a new variable. - Determine which factors can equal zero for real \(x\). - A real square is never negative.

1. Factor out \(3\): \(f(x) = 3(x^4 + 2x^2 - 3)\). 2. Let \(u = x^2\). Then \(u^2 + 2u - 3 = 0\). 3. Factor: \((u - 1)(u + 3) = 0\), so \(u = 1\) or \(u = -3\). 4. Substitute back: \(f(x) = 3(x^2 - 1)(x^2 + 3) = 3(x - 1)(x + 1)(x^2 + 3)\). 5. The linear factors give the real zeros \(x = -1\) and \(x = 1\). 6. Since \(x^2 + 3 > 0\) for every real \(x\), that factor gives no additional real zeros.

a) \(f(x) = 3(x - 1)(x + 1)(x^2 + 3)\) b) Real zeros: \(x = -1, 1\). The factor \(x^2 + 3\) has no real zeros.

- Solutions of an equation with only even powers occur in opposite pairs. - Use the solutions to write factors. - Pair opposite roots to form expressions of the form \(x^2 - c\). - Expand and compare coefficients.

1. Because the equation contains only even powers of \(x\), the opposite values \(-\sqrt{2}\) and \(-5\) are also solutions. 2. Write the equation in factored form: \((x - \sqrt{2})(x + \sqrt{2})(x - 5)(x + 5) = 0\). 3. Combine conjugate factors: \((x^2 - 2)(x^2 - 25) = 0\). 4. Expand: \(x^4 - 27x^2 + 50 = 0\). 5. Therefore, \(a = -27\) and \(b = 50\).

\(a = -27\), \(b = 50\); the equation is \(x^4 - 27x^2 + 50 = 0\).

- Convert the known value of \(x\) into a value of \(u\). - Build a quadratic equation from its two solutions. - Compare the expanded equation with \(u^2 + pu + q = 0\). - Substitute back for both values of \(u\).

1. Since \(x = 3\) is a solution, one solution of the quadratic equation in \(u\) is \(u = 3^2 = 9\). 2. The two solutions in \(u\) are \(9\) and \(1\), so \((u - 9)(u - 1) = 0\). 3. Expand: \(u^2 - 10u + 9 = 0\). Therefore, \(p = -10\) and \(q = 9\). 4. Substitute back. From \(x^2 = 9\), \(x = \pm 3\). From \(x^2 = 1\), \(x = \pm 1\).

\(p = -10\), \(q = 9\); real solutions: \(\{-3, -1, 1, 3\}\)

- Use each solution to form a factor. - Pair opposite roots and apply the difference-of-squares pattern. - Expand the two quadratic factors.

1. Use the given solutions to write factors: \((x - \sqrt{5})(x + \sqrt{5})(x - \sqrt{2})(x + \sqrt{2}) = 0\). 2. Pair opposite roots: \((x^2 - 5)(x^2 - 2) = 0\). 3. Expand: \(x^4 - 7x^2 + 10 = 0\).

\(x^4 - 7x^2 + 10 = 0\)

- Substitute a variable for \(x^{\frac{1}{3}}\). - Rewrite \(x^{\frac{2}{3}}\) as the square of that substituted variable. - Solve the quadratic equation, then substitute back.

1. Let \(u=x^{\frac{1}{3}}\). Then \(x^{\frac{2}{3}}=u^2\), so the equation becomes \(u^2-4u+3=0\). 2. Factor: \((u-1)(u-3)=0\). Thus \(u=1\) or \(u=3\). 3. Substitute back. If \(x^{\frac{1}{3}}=1\), then \(x=1\). If \(x^{\frac{1}{3}}=3\), then \(x=27\).

\(\{1,27\}\)

- Substitute for the entire radical expression. - Express \(x^2+2x\) in terms of the substituted variable. - Reject negative values for a variable defined as a square root. - Substitute back to solve for \(x\).

1. Let \(u=\sqrt{x^2+2x+8}\), so \(u\ge0\) and \(u^2=x^2+2x+8\). Therefore, \(x^2+2x=u^2-8\). 2. Substitute: \(u^2-8+u=12\), so \(u^2+u-20=0\). 3. Factor: \((u+5)(u-4)=0\). Since \(u\ge0\), use \(u=4\). 4. Substitute back: \(\sqrt{x^2+2x+8}=4\). Squaring gives \(x^2+2x-8=0\). 5. Factor: \((x+4)(x-2)=0\), so \(x=-4\) or \(x=2\). Both satisfy the original equation.

\(\{-4,2\}\)

- Relate \(2x^2-4x\) to the expression under the radical. - A variable defined as a square root must be nonnegative. - After solving for \(u\), substitute back and solve the resulting quadratic equation.

1. From \(u^2=x^2-2x+6\), it follows that \(x^2-2x=u^2-6\). Since \(2x^2-4x=2(x^2-2x)\), substitution gives \(2(u^2-6)+3u=15\), or \(2u^2+3u-27=0\). 2. Solve the quadratic equation: \(u=\frac{-3\pm15}{4}\), so \(u=3\) or \(u=-4.5\). Because \(u\) is a square root, only \(u=3\) is valid. 3. Substitute back: \(\sqrt{x^2-2x+6}=3\). Squaring gives \(x^2-2x-3=0\). 4. Factor: \((x-3)(x+1)=0\). Thus \(x=3\) or \(x=-1\), and both satisfy the original equation.

a) Substitution gives \(2u^2+3u-27=0\). b) \(\{-1,3\}\)

- Substitute the expression for \(y\) into the other equation. - Look for an equation in quadratic form. - Include both square roots when solving for \(x\). - Use \(y = x^2 - 5\) to find the matching y-values.

1. Substitute \(y = x^2 - 5\) into the second equation: \((x^2 - 5)^2 + 2x^2 = 10\). 2. Expand and simplify: \(x^4 - 8x^2 + 15 = 0\). 3. Let \(u = x^2\). Then \(u^2 - 8u + 15 = 0\). 4. Factor: \((u - 3)(u - 5) = 0\), so \(u = 3\) or \(u = 5\). 5. If \(x^2 = 5\), then \(x = \pm\sqrt{5}\) and \(y = 0\). 6. If \(x^2 = 3\), then \(x = \pm\sqrt{3}\) and \(y = -2\). 7. Therefore, there are four solutions.

\((\sqrt{5}, 0)\), \((-\sqrt{5}, 0)\), \((\sqrt{3}, -2)\), and \((-\sqrt{3}, -2)\); 4 solutions

- Eliminate one variable using \(xy = 4\). - Clear the denominator to obtain a polynomial equation. - Replace \(x^2\) with a new variable. - Substitute each x-value back to find its matching y-value.

1. Since \(xy = 4\), \(x \neq 0\), and \(y = \frac{4}{x}\). 2. Substitute into the first equation: \(x^2 + \left(\frac{4}{x}\right)^2 = 17\). 3. Multiply by \(x^2\): \(x^4 - 17x^2 + 16 = 0\). 4. Let \(u = x^2\). Then \(u^2 - 17u + 16 = 0\). 5. Factor: \((u - 1)(u - 16) = 0\), so \(u = 1\) or \(u = 16\). 6. Thus, \(x = \pm 1\) or \(x = \pm 4\). 7. Use \(y = \frac{4}{x}\) to obtain the matching values of \(y\).

\((4, 1)\), \((-4, -1)\), \((1, 4)\), and \((-1, -4)\)

- Replace \(x^2\) with a new variable. - Remember that a positive value of \(x^2\) gives two opposite x-values. - Multiply the four solutions directly. - Connect y-axis symmetry with zeros occurring in opposite pairs.

1. Let \(u = x^2\). Then \(u^2 - 10u + 9 = 0\). 2. Factor: \((u - 1)(u - 9) = 0\), so \(u = 1\) or \(u = 9\). 3. Substitute back to get the real solution set \(\{-3, -1, 1, 3\}\). 4. The product is \((-3) \cdot (-1) \cdot 1 \cdot 3 = 9\), equal to the constant term. 5. Because \(f\) is even, every nonzero zero \(a\) occurs with the zero \(-a\). Each pair has sum \(0\), so the sum of all four zeros is \(0\).

1. \(\{-3, -1, 1, 3\}\) 2. \((-3) \cdot (-1) \cdot 1 \cdot 3 = 9\) 3. The zeros occur in opposite pairs \(\pm a\), so their total sum is \(0\).

- Look for an expression that appears more than once. - Replace that entire expression with one variable. - During back-substitution, determine whether each equation of the form \(x^2 = a\) has real solutions. - Check each solution in the original equation.

1. Let \(u = x^2 - 1\). The equation becomes \(u^2 - 7u - 8 = 0\). 2. Factor: \((u - 8)(u + 1) = 0\), so \(u = 8\) or \(u = -1\). 3. If \(u = 8\), then \(x^2 - 1 = 8\), so \(x^2 = 9\) and \(x = \pm 3\). 4. If \(u = -1\), then \(x^2 - 1 = -1\), so \(x^2 = 0\) and \(x = 0\).

\(x \in \{-3, 0, 3\}\)

- Set the function value equal to zero. - Compare the exponents \(6\) and \(3\) and look for a repeated-power structure that can turn the equation into a quadratic. - Solve the resulting quadratic equation first. - Convert the resulting power equations back to real values of \(x\).

1. Set the function equal to zero: \(x^6 + 26x^3 - 27 = 0\). 2. Substitute \(u = x^3\). Then \(x^6 = u^2\), so the equation becomes \(u^2 + 26u - 27 = 0\). 3. Factor the quadratic: \(u^2 + 26u - 27 = (u - 1)(u + 27)\). Thus, \(u = 1\) or \(u = -27\). 4. Substitute back. From \(x^3 = 1\), \(x = 1\). From \(x^3 = -27\), \(x = -3\). 5. Therefore, the real zeros are \(x = -3\) and \(x = 1\).

The real zeros are \(x = -3\) and \(x = 1\).

- Look for a substitution that turns the equation into a quadratic equation. - After solving for the substituted variable, replace it with the original expression. - Each equation of the form \(x^3=a\) has exactly one real solution.

1. Because the exponents are \(6\) and \(3\), let \(z=x^3\). Then \(x^6=z^2\). 2. The equation \(h(x)=0\) becomes \(z^2-7z-8=0\). 3. Factor: \(z^2-7z-8=(z-8)(z+1)\). Thus, \(z=8\) or \(z=-1\). 4. Substitute back. From \(x^3=8\), \(x=2\). From \(x^3=-1\), \(x=-1\).

The real zeros are \(x=-1\) and \(x=2\).

- Replace \(x^2\) with a new variable to obtain a quadratic equation. - Each positive value found for the new variable gives two possible x-values. - Count the points where the curve crosses the x-axis.

1. Let \(u=x^2\). Then \(u^2-5u+4=0\). 2. Factor: \((u-4)(u-1)=0\), so \(u=4\) or \(u=1\). 3. Back-substitute. From \(x^2=4\), \(x=\pm2\). From \(x^2=1\), \(x=\pm1\). 4. The function has four distinct real zeros, so its graph has four x-intercepts: \((-2, 0)\), \((-1, 0)\), \((1, 0)\), and \((2, 0)\).

The zeros are \(x=-2\), \(x=-1\), \(x=1\), and \(x=2\). Therefore, the graph has four x-intercepts.

- Use the repeated square-root expression as the new variable. - Record the sign restriction that comes with that substitution. - Check each quadratic root against the substitution restriction before converting back to \(x\).

1. Let \(u=\sqrt{x}\), so \(u\ge0\) and \(x=u^2\). 2. The equation becomes \(u^2+2u-8=0\). 3. Factor: \((u+4)(u-2)=0\), so \(u=-4\) or \(u=2\). 4. The substitution requires \(u\ge0\), so reject \(u=-4\). 5. From \(u=2\), \(x=4\). Substitution verifies the original equation.

\(x=4\)

- Which repeated expression becomes one new variable? - Factor the quadratic in that new variable before returning to \(x\). - Remember both signs when undoing a square.

1. Substitute to get \(u^2-5u+4=0=(u-1)(u-4)\), so \(u=1\) or \(u=4\). 2. From \((x-2)^2=1\), get \(x=1,3\). From \((x-2)^2=4\), get \(x=0,4\). 3. Each value satisfies its squared branch and therefore the original equation.

With \(u=(x-2)^2\), \((u-1)(u-4)=0\); thus \(x\in\{0,1,3,4\}\).

- Which power can replace the middle term and whose square makes the leading term? - After finding the substituted values, undo a cube rather than a square.

1. Substitution gives \(u^2-7u+6=(u-1)(u-6)=0\). 2. Thus \(x^3=1\) or \(x^3=6\), giving the real solutions \(x=1\) and \(x=\sqrt[3]{6}\). 3. Each solution has a cube equal to one of the substituted roots and checks in the original equation.

\(u^2-7u+6=(u-1)(u-6)=0\) with \(u=x^3\); hence \(x=1\) or \(x=\sqrt[3]{6}\).

- An x-intercept is a point where the function value is zero. - What values of \(u\) would make the rewritten equation equal zero? - Remember the negative as well as positive square roots.

1. The graph crosses the x-axis at \(x=-2,-1,1,2\). 2. Substitute \(u=x^2\) to get \(u^2-5u+4=(u-1)(u-4)=0\), so \(u=1\) or \(u=4\). 3. Back-substitute to get \(x=\pm1\) and \(x=\pm2\), agreeing with the graph.

The x-intercepts are \((-2,0)\), \((-1,0)\), \((1,0)\), and \((2,0)\). With \(u=x^2\), \((u-1)(u-4)=0\), so \(u=1\) or \(u=4\), giving \(x=\pm1\) or \(x=\pm2\).

- Replace one rational expression with a new variable. - Notice that the two fractions in the first equation are reciprocals. - Convert each value of the new variable into a linear relationship between \(x\) and \(y\). - Verify that neither original denominator is zero.

1. The denominators require \(x\ne y\) and \(x\ne-y\). Let \(u=\frac{x+y}{x-y}\). Then the first equation becomes \(u+\frac{1}{u}=\frac{10}{3}\). 2. Multiply by \(3u\): \(3u^2-10u+3=0\). Factor: \((3u-1)(u-3)=0\), so \(u=3\) or \(u=\frac{1}{3}\). 3. If \(u=3\), then \(x+y=3(x-y)\), so \(x=2y\). Substituting into \(2x+y=7\) gives \(5y=7\), so \((x, y)=\left(\frac{14}{5},\frac{7}{5}\right)\). 4. If \(u=\frac{1}{3}\), then \(3(x+y)=x-y\), so \(x=-2y\). Substituting into \(2x+y=7\) gives \(-3y=7\), so \((x, y)=\left(\frac{14}{3},-\frac{7}{3}\right)\). 5. Both ordered pairs satisfy the domain restrictions.

Solution set: \(\left\{\left(\frac{14}{5},\frac{7}{5}\right),\left(\frac{14}{3},-\frac{7}{3}\right)\right\}\).

- Express \(y^2\) in terms of \(x^2\). - Substitute into the fourth-power equation. - Treat \(x^2\) as a new variable. - For each value of \(x^2\), include all possible signs of \(x\) and \(y\).

1. From the second equation, \(y^2 = 5 - x^2\). 2. Substitute into the first equation: \(x^4 + (5 - x^2)^2 = 17\). 3. Expand and simplify: \(2x^4 - 10x^2 + 8 = 0\), or \(x^4 - 5x^2 + 4 = 0\). 4. Let \(u = x^2\). Then \(u^2 - 5u + 4 = 0\), so \(u = 1\) or \(u = 4\). 5. If \(x^2 = 4\), then \(x = \pm 2\) and \(y^2 = 1\), so \(y = \pm 1\). 6. If \(x^2 = 1\), then \(x = \pm 1\) and \(y^2 = 4\), so \(y = \pm 2\). 7. These choices produce eight ordered pairs, and each ordered pair is one intersection point.

1. \((2, 1)\), \((2, -1)\), \((-2, 1)\), \((-2, -1)\), \((1, 2)\), \((1, -2)\), \((-1, 2)\), \((-1, -2)\) 2. 8 intersection points

- Substitute a new variable for one square-root ratio. - The two square-root ratios are reciprocals. - Solve the resulting equation in quadratic form, then use the sum equation.

1. Let \(t = \sqrt{\frac{x}{y}}\). The domain and the positive sum imply \(t > 0\), and \(\sqrt{\frac{y}{x}} = \frac{1}{t}\). 2. The first equation becomes \(t + \frac{1}{t} = \frac{10}{3}\). Multiply by \(3t\): \(3t^2 - 10t + 3 = 0\). 3. Factor: \((3t - 1)(t - 3) = 0\). Thus, \(t = 3\) or \(t = \frac{1}{3}\). 4. If \(t = 3\), then \(x = 9y\). With \(x + y = 30\), this gives \((x, y) = (27, 3)\). 5. If \(t = \frac{1}{3}\), then \(y = 9x\). With \(x + y = 30\), this gives \((x, y) = (3, 27)\).

\(\{(27, 3), (3, 27)\}\)

- Replace \(\frac{x}{y}\) with a new variable. - Use the fact that \(\frac{y}{x}\) is the reciprocal of that variable. - After solving for the substitution variable, return to the original variables. - Account for both signs when solving a squared equation.

1. Let \(u=\frac{x}{y}\), where \(u\ne0\). Then the first equation becomes \(u+\frac{1}{u}=\frac{5}{2}\). 2. Multiply by \(2u\): \(2u^2-5u+2=0\). 3. Factor: \((2u-1)(u-2)=0\), so \(u=2\) or \(u=\frac{1}{2}\). 4. If \(u=2\), then \(x=2y\). Substituting into \(x^2+y^2=20\) gives \(5y^2=20\), so \(y=\pm2\). This yields \((4, 2)\) and \((-4, -2)\). 5. If \(u=\frac{1}{2}\), then \(y=2x\). Substitution gives \(5x^2=20\), so \(x=\pm2\). This yields \((2, 4)\) and \((-2, -4)\). 6. All four pairs satisfy the nonzero restrictions.

Solution set: \(\{(-4, -2),(-2, -4),(2, 4),(4, 2)\}\).

- Track the signs carefully when solving for the substitution variable. - Substitute each value of the new variable back into the second equation. - A negative value for a real square produces no real solutions.

1. Let \(u=\frac{x}{y}\), where \(u\ne0\). The first equation becomes \(u-\frac{1}{u}=\frac{7}{12}\). 2. Multiply by \(12u\): \(12u^2-7u-12=0\). 3. Factor: \((3u-4)(4u+3)=0\), so \(u=\frac{4}{3}\) or \(u=-\frac{3}{4}\). 4. If \(u=\frac{4}{3}\), then \(x=\frac{4}{3}y\). Substitution into \(x^2-y^2=7\) gives \(\frac{7}{9}y^2=7\), so \(y=\pm3\). This yields \((4, 3)\) and \((-4, -3)\). 5. If \(u=-\frac{3}{4}\), then \(x=-\frac{3}{4}y\). Substitution gives \(-\frac{7}{16}y^2=7\), or \(y^2=-16\), which has no real solutions. 6. Therefore, the system has only two real solutions. The claim is false because a valid substitution value need not produce real values of the original variables.

a) Solution set: \(\{(-4, -3),(4, 3)\}\). b) The second substitution case leads to \(y^2=-16\), so it produces no real ordered pairs. The system has two, not four, real solutions.

- Pair the factors so both products contain \(x^2 + x\). - Substitute a new variable for the repeated quadratic expression. - Solve in the new variable, then solve each resulting quadratic in \(x\).

1. Group the factors as \((x - 3)(x + 4)\) and \((x - 1)(x + 2)\): \((x^2 + x - 12)(x^2 + x - 2) = 144\). 2. Let \(u = x^2 + x\). Then \((u - 12)(u - 2) = 144\). 3. Expand and simplify: \(u^2 - 14u - 120 = 0\). 4. Factor: \((u - 20)(u + 6) = 0\). Thus, \(u = 20\) or \(u = -6\). 5. If \(u = 20\), then \(x^2 + x - 20 = 0\), so \((x + 5)(x - 4) = 0\). This gives \(x = -5\) or \(x = 4\). 6. If \(u = -6\), then \(x^2 + x + 6 = 0\). Its discriminant is \(1 - 24 = -23\), so it has no real solutions.

\(x \in \{-5, 4\}\)

- A positive value of \(x^2\) gives two real x-values, while \(x^2 = 0\) gives one. - Determine what combination of values for \(x^2\) produces exactly three solutions. - If \(u = 0\) is a solution, substitute it into the quadratic equation in \(u\). - Check that the other value of \(u\) is positive.

1. For a), factor: \(x^2(x^2 - 25) = 0\). 2. The first factor gives \(x = 0\), and the second gives \(x = \pm 5\). Thus, there are exactly three distinct real solutions. 3. For b), let \(u = x^2\). Then \(u^2 - 12u + k = 0\). 4. To produce exactly three real x-values, one value of \(u\) must be \(0\), giving one x-value, and the other must be positive, giving two x-values. 5. Substituting \(u = 0\) gives \(k = 0\). 6. With \(k = 0\), the equation is \(u(u - 12) = 0\), so \(u = 0\) or \(u = 12\). 7. Therefore, \(x = 0\) or \(x = \pm\sqrt{12} = \pm 2\sqrt{3}\).

a) \(\{-5, 0, 5\}\) b) \(k = 0\); solutions: \(\{-2\sqrt{3}, 0, 2\sqrt{3}\}\)

- Replace \(x^2\) with a new variable. - Factor the quadratic using its sum and product structure. - Substitute back to obtain \(x = \pm\sqrt{u}\). - Check when the listed solutions coincide.

1. Let \(u = x^2\). Then \(u^2 - (k^2 + 9)u + 9k^2 = 0\). 2. Factor the quadratic: \((u - 9)(u - k^2) = 0\). 3. Thus, \(x^2 = 9\) or \(x^2 = k^2\), giving \(x = \pm 3\) or \(x = \pm k\). 4. When \(k \notin \{0, -3, 3\}\), these are four distinct solutions. 5. When \(k = 0\), the distinct solutions are \(-3, 0, 3\). 6. When \(k = 3\) or \(k = -3\), the distinct solutions are only \(-3\) and \(3\).

If \(k \notin \{0, -3, 3\}\): \(\{-3, 3, -k, k\}\) If \(k = 0\): \(\{-3, 0, 3\}\) If \(k = \pm 3\): \(\{-3, 3\}\)

- Replace \(x^2\) with a new variable. - Look for a factorization involving \(u - a^2\). - Use \(a > 0\) when taking square roots. - Check when the two pairs of solutions coincide.

1. Let \(u = x^2\). Then \(a^2u^2 - (a^4 + 1)u + a^2 = 0\). 2. Factor the quadratic in \(u\): \((u - a^2)(a^2u - 1) = 0\). 3. Therefore, \(u = a^2\) or \(u = \frac{1}{a^2}\). 4. Substitute back. From \(x^2 = a^2\), \(x = \pm a\). From \(x^2 = \frac{1}{a^2}\), \(x = \pm\frac{1}{a}\). 5. If \(a \neq 1\), these are four distinct solutions. If \(a = 1\), the two pairs coincide, leaving \(x = \pm 1\).

If \(a \neq 1\): \(\left\{-a, a, -\frac{1}{a}, \frac{1}{a}\right\}\) If \(a = 1\): \(\{-1, 1\}\)

- If \(\sqrt{5}\) is a solution, use the equation's even-power structure to identify another solution. - Relate the four x-solutions to the two roots of the substituted equation. - Use Vieta's formulas for the sum and product of the substituted roots. - Determine when \(x^2 = u\) has no real solution.

1. Because the equation contains only even powers, \(-\sqrt{5}\) is also a solution. 2. Let the other pair of solutions be \(r\) and \(-r\). Their total product is \((\sqrt{5})(-\sqrt{5})(r)(-r) = 5r^2\). 3. Since the product is \(20\), \(5r^2 = 20\), so \(r^2 = 4\) and \(r = 2\). The other solutions are \(2\) and \(-2\). 4. After substituting \(u = x^2\), the two roots are \(u = 5\) and \(u = 4\). 5. By Vieta's formulas, \(p = -(5 + 4) = -9\) and \(q = 5 \cdot 4 = 20\). 6. If \(q < 0\), the two roots of \(u^2 + pu + q = 0\) have opposite signs because their product is \(q\). Only the positive root can equal \(x^2\), and it gives exactly two real values \(x = \pm\sqrt{u}\).

1. \(-\sqrt{5}\), \(-2\), and \(2\) 2. \(p = -9\), \(q = 20\) 3. When \(q < 0\), the substituted roots have opposite signs. Only the positive root gives real x-values, producing exactly two real solutions.

- Factor \(x^2-y^2\). - Replace \(x-y\) and \(x+y\) with new variables. - Use the product equation to express one new variable in terms of the other. - Convert the substitution values back to \(x\) and \(y\).

1. Factor the second equation: \((x-y)(x+y)=8\). 2. Let \(u=x-y\) and \(v=x+y\). Then \(uv=8\), so \(v=\frac{8}{u}\). 3. Substitute into the first equation: \(\frac{1}{u}+\frac{1}{v}=\frac{3}{4}\), so \(\frac{1}{u}+\frac{u}{8}=\frac{3}{4}\). 4. Multiply by \(8u\): \(8+u^2=6u\), giving \(u^2-6u+8=0\). 5. Factor: \((u-2)(u-4)=0\), so \(u=2\) or \(u=4\). The corresponding values of \(v\) are \(4\) and \(2\). 6. Solve \(x-y=2\), \(x+y=4\) to obtain \((x, y)=(3, 1)\). Solve \(x-y=4\), \(x+y=2\) to obtain \((x, y)=(3, -1)\). 7. Both pairs keep the original denominators nonzero.

Solution set: \(\{(3, -1),(3, 1)\}\).

- Substitute new variables for \(x + y\) and \(xy\). - Use the given sum and product to form a quadratic equation. - Back-substitute each possible sum-product pair.

1. Let \(u = x + y\) and \(v = xy\). Then \(u + v = 11\) and \(uv = 30\). 2. The values \(u\) and \(v\) satisfy \(t^2 - 11t + 30 = 0\). Factor: \((t - 5)(t - 6) = 0\). 3. Thus, either \((u, v) = (5, 6)\) or \((u, v) = (6, 5)\). 4. If \(x + y = 5\) and \(xy = 6\), then \(x\) and \(y\) are \(2\) and \(3\), giving \((2, 3)\) and \((3, 2)\). 5. If \(x + y = 6\) and \(xy = 5\), then \(x\) and \(y\) are \(1\) and \(5\), giving \((1, 5)\) and \((5, 1)\).

\(\{(1, 5), (5, 1), (2, 3), (3, 2)\}\)

- Look for a squared expression that would produce both \(x^2\) and \(\frac{1}{x^2}\). - After identifying a repeated reciprocal expression, rewrite the equation in terms of that one quantity. - Each value of the substituted quantity leads to a second equation in \(x\). - Keep the original nonzero restriction through the back-substitution.

1. Notice that \(\left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}\). Let \(u=x+\frac{1}{x}\). 2. Then \(x^2+\frac{1}{x^2}=u^2-2\), so the equation becomes \(u^2-5u+6=0\). 3. Factor: \((u-2)(u-3)=0\), so \(u=2\) or \(u=3\). 4. If \(u=2\), then \(x+\frac{1}{x}=2\). Multiplying by \(x\) gives \(x^2-2x+1=0\), so \(x=1\). 5. If \(u=3\), then \(x+\frac{1}{x}=3\). Multiplying by \(x\) gives \(x^2-3x+1=0\), so \(x=\frac{3\pm\sqrt{5}}{2}\). 6. All three values are real and nonzero, so all satisfy the original domain restriction.

\(x=1\), \(x=\frac{3-\sqrt{5}}{2}\), or \(x=\frac{3+\sqrt{5}}{2}\)

- Combine the equations so the constants cancel. - Divide the homogeneous quadratic by \(y^2\) only after justifying that \(y \ne 0\). - Solve the equation in \(t = \frac{x}{y}\), then substitute each ratio into an original equation.

1. Multiply the first equation by \(5\) and the second by \(7\), then add: \(5x^2 - 19xy + 12y^2 = 0\). 2. The second original equation shows that \(y \ne 0\). Divide by \(y^2\) and let \(t = \frac{x}{y}\): \(5t^2 - 19t + 12 = 0\). 3. Factor: \((5t - 4)(t - 3) = 0\). Thus, \(t = \frac{4}{5}\) or \(t = 3\). 4. If \(x = 3y\), substitute into \(y^2 - 2xy = -15\): \(-5y^2 = -15\), so \(y = \pm\sqrt{3}\). This gives \((3\sqrt{3}, \sqrt{3})\) and \((-3\sqrt{3}, -\sqrt{3})\). 5. If \(x = \frac{4}{5}y\), then \(-\frac{3}{5}y^2 = -15\), so \(y = \pm 5\). This gives \((4, 5)\) and \((-4, -5)\).

\(\{(3\sqrt{3}, \sqrt{3}), (-3\sqrt{3}, -\sqrt{3}), (4, 5), (-4, -5)\}\)