Free Math Worksheets

- Move the constant term to the other side. - Add the value that turns the left side into a perfect-square trinomial. - Use both square roots when solving the resulting equation.

1. Move the constant term: \(x^2 + 2x = 3\). 2. Add \(1\) to both sides: \(x^2 + 2x + 1 = 4\). 3. Rewrite the left side: \((x + 1)^2 = 4\). 4. Take both square roots: \(x + 1 = \pm 2\). 5. Therefore, \(x = 1\) or \(x = -3\).

\(x = 1\) or \(x = -3\)

- Move the constant term to the other side. - Complete the square on the left side. - Keep the solutions in exact radical form and use both square roots.

1. Move the constant term: \(x^2 - 4x = -1\). 2. Add \(4\) to both sides: \(x^2 - 4x + 4 = 3\). 3. Rewrite the left side: \((x - 2)^2 = 3\). 4. Take both square roots: \(x - 2 = \pm\sqrt{3}\). 5. Therefore, \(x = 2 + \sqrt{3}\) or \(x = 2 - \sqrt{3}\).

\(x = 2 + \sqrt{3}\) or \(x = 2 - \sqrt{3}\)

- Move the constant term to the other side. - Use half the coefficient of \(x\) to determine what to add to both sides. - Simplify the radical after taking both square roots.

1. Move the constant term: \(x^2 + 10x = -17\). 2. Add \(25\) to both sides: \(x^2 + 10x + 25 = 8\). 3. Rewrite the left side: \((x + 5)^2 = 8\). 4. Take both square roots and simplify: \(x + 5 = \pm\sqrt{8} = \pm 2\sqrt{2}\). 5. Therefore, \(x = -5 + 2\sqrt{2}\) or \(x = -5 - 2\sqrt{2}\).

\(x = -5 + 2\sqrt{2}\) or \(x = -5 - 2\sqrt{2}\)

- Do the first two expressions match a perfect-square trinomial pattern? - For c), what number completes the square for \(x^2 - 2x\)? - How can you rewrite the constant term so that part of it completes the square?

1. For a), \(x^2 - 12x + 36 = x^2 - 2 \cdot 6x + 6^2 = (x - 6)^2\). Thus, \(f(x) = (x - 6)^2\), and the vertex is \((6, 0)\). 2. For b), \(6.25 = 2.5^2\), so \(x^2 + 5x + 6.25 = x^2 + 2 \cdot 2.5x + 2.5^2 = (x + 2.5)^2\). Thus, \(g(x) = (x + 2.5)^2\), and the vertex is \((-2.5, 0)\). 3. For c), complete the square: \(x^2 - 2x + 5 = x^2 - 2x + 1 + 4 = (x - 1)^2 + 4\). Thus, \(h(x) = (x - 1)^2 + 4\), and the vertex is \((1, 4)\).

a) \(f(x) = (x - 6)^2\); vertex \((6, 0)\) b) \(g(x) = (x + 2.5)^2\); vertex \((-2.5, 0)\) c) \(h(x) = (x - 1)^2 + 4\); vertex \((1, 4)\)

- Relate the linear term to a squared binomial. - What must the vertex's y-coordinate be when the vertex lies on the x-axis? - For \(x^2+px+q\), how is \(q\) related to half of \(p\)?

1. A monic quadratic with its vertex on the x-axis is a perfect square. For \(x^2+px+q\), the required constant is \(q=\left(\frac{p}{2}\right)^2\). 2. For \(f\), \(\left(\frac{12}{2}\right)^2=36\). Thus, \(f(x)=x^2+12x+36=(x+6)^2\), with vertex \((-6, 0)\). 3. For \(g\), \(\left(\frac{-18}{2}\right)^2=81\). Thus, \(g(x)=x^2-18x+81=(x-9)^2\), with vertex \((9, 0)\). 4. For \(h\), \(\left(\frac{7}{2}\right)^2=12.25\). Thus, \(h(x)=x^2+7x+12.25=(x+3.5)^2\), with vertex \((-3.5, 0)\).

a) \(f(x)=x^2+12x+36\); vertex \((-6, 0)\) b) \(g(x)=x^2-18x+81\); vertex \((9, 0)\) c) \(h(x)=x^2+7x+12.25\); vertex \((-3.5, 0)\)

- What number completes the square for \(x^2 + 10x\)? - How does the sign inside the squared binomial determine the vertex's \(x\)-coordinate? - What does the sign of the leading coefficient tell you about the opening of the parabola?

1. Half of \(10\) is \(5\), and \(5^2 = 25\). 2. Add and subtract \(25\): \(f(x) = x^2 + 10x + 25 - 25 + 7\). 3. Rewrite the perfect-square trinomial: \(f(x) = (x + 5)^2 - 18\). 4. The vertex is \((-5, -18)\). Since the coefficient of the squared term is positive, the parabola opens upward, so the vertex is a minimum.

The vertex form is \(f(x) = (x + 5)^2 - 18\). The vertex is \((-5, -18)\), and it is a minimum.

- What number completes the square for \(x^2 + 4x\)? - Remember to subtract the same value you add. - How do you read the vertex from \(a(x - h)^2 + k\)?

1. Complete the square: \(x^2 + 4x + 7 = x^2 + 4x + 4 - 4 + 7\). 2. Rewrite and simplify: \(f(x) = (x + 2)^2 + 3\). 3. Therefore, the vertex is \((-2, 3)\).

\(f(x) = (x + 2)^2 + 3\); vertex \((-2, 3)\)

- Move the constant term to the other side. - Add the value needed to form a perfect-square trinomial. - Use both square roots and keep the answer in exact form.

1. Move the constant term: \(x^2 + 2x = 4\). 2. Add \(1\) to both sides: \(x^2 + 2x + 1 = 5\). 3. Rewrite the left side: \((x + 1)^2 = 5\). 4. Take both square roots: \(x + 1 = \pm\sqrt{5}\). 5. Therefore, \(x = -1 + \sqrt{5}\) or \(x = -1 - \sqrt{5}\).

\(x = -1 + \sqrt{5}\) or \(x = -1 - \sqrt{5}\)

- Move the constant term before completing the square. - Add the square of half the coefficient of \(x\). - After taking square roots, simplify the radical.

1. Move the constant term: \(x^2 + 4x = 4\). 2. Add \(4\) to both sides: \(x^2 + 4x + 4 = 8\). 3. Rewrite the left side: \((x + 2)^2 = 8\). 4. Take both square roots: \(x + 2 = \pm\sqrt{8} = \pm 2\sqrt{2}\). 5. Therefore, \(x = -2 + 2\sqrt{2}\) or \(x = -2 - 2\sqrt{2}\).

\(x = -2 + 2\sqrt{2}\) or \(x = -2 - 2\sqrt{2}\)

- Write each equation in the form \(x^2 + bx = c\). - Add the square of half the coefficient of \(x\) to both sides. - Use both square roots after forming a perfect square.

1. For equation 1, write \(x^2 + 10x = 24\). Add \(25\) to both sides: \((x + 5)^2 = 49\). 2. Take both square roots: \(x + 5 = \pm 7\). Thus, \(x = 2\) or \(x = -12\). 3. For equation 2, add \(64\) to both sides: \((x - 8)^2 = 49\). 4. Take both square roots: \(x - 8 = \pm 7\). Thus, \(x = 15\) or \(x = 1\).

1) \(x \in \{-12, 2\}\) 2) \(x \in \{1, 15\}\)

- Move the constant term to the other side. - Add the square of half the coefficient of \(x\) to both sides. - Use both square roots.

1. For part a, write \(x^2 + 4x = 21\). Add \(4\) to both sides: \((x + 2)^2 = 25\). 2. Take both square roots: \(x + 2 = \pm 5\). Thus, \(x = 3\) or \(x = -7\). 3. For part b, write \(x^2 - 6x = -5\). Add \(9\) to both sides: \((x - 3)^2 = 4\). 4. Take both square roots: \(x - 3 = \pm 2\). Thus, \(x = 5\) or \(x = 1\).

a) \(x \in \{-7, 3\}\) b) \(x \in \{1, 5\}\)

- First isolate the terms containing \(x\) and make the coefficient of \(x^2\) equal to \(1\). - Add the value needed to create a perfect-square trinomial. - Remember to use both the positive and negative square roots.

1. Move the constant term and divide by \(4\): \(4x^2 - 16x = 9\), so \(x^2 - 4x = \frac{9}{4}\). 2. Add \(4\) to both sides: \(x^2 - 4x + 4 = \frac{9}{4} + 4\). 3. Rewrite and simplify: \((x - 2)^2 = \frac{25}{4}\). 4. Take both square roots: \(x - 2 = \pm\frac{5}{2}\). 5. Therefore, \(x = \frac{9}{2}\) or \(x = -\frac{1}{2}\).

\(x = \frac{9}{2}\) or \(x = -\frac{1}{2}\)

- Make the coefficient of \(x^2\) equal to \(1\) before completing the square. - Add the value needed to form a perfect-square trinomial. - Keep the final solutions in exact radical form.

1. Move the constant term and divide by \(4\): \(4x^2 - 24x = -29\), so \(x^2 - 6x = -\frac{29}{4}\). 2. Add \(9\) to both sides: \(x^2 - 6x + 9 = -\frac{29}{4} + 9\). 3. Rewrite and simplify: \((x - 3)^2 = \frac{7}{4}\). 4. Take both square roots: \(x - 3 = \pm\frac{\sqrt{7}}{2}\). 5. Therefore, \(x = 3 + \frac{\sqrt{7}}{2}\) or \(x = 3 - \frac{\sqrt{7}}{2}\).

\(x = 3 + \frac{\sqrt{7}}{2}\) or \(x = 3 - \frac{\sqrt{7}}{2}\)

- Before completing the square, what should you do with the coefficient of \(x^2\)? - How do you find the number needed to complete the square? - In vertex form, how does the sign inside the parentheses determine the horizontal shift? - How does a coefficient outside the squared expression affect the parabola?

1. Factor \(2\) from the quadratic and linear terms: \(f(x)=2(x^2-6x)+10\). 2. Complete the square inside the parentheses using \(\left(\frac{-6}{2}\right)^2=9\): \(f(x)=2(x^2-6x+9-9)+10\). 3. Rewrite the perfect-square trinomial: \(f(x)=2((x-3)^2-9)+10\). 4. Distribute and simplify: \(f(x)=2(x-3)^2-18+10=2(x-3)^2-8\). 5. The vertex is \((3, -8)\). 6. Relative to \(y=x^2\), the graph is vertically stretched by a factor of \(2\), shifted \(3\) units right, and shifted \(8\) units down.

The vertex form is \(f(x)=2(x-3)^2-8\), and the vertex is \((3, -8)\). The graph is vertically stretched by a factor of \(2\), shifted \(3\) units right, and shifted \(8\) units down.

- How can completing the square reveal the vertex of \(g\)? - Compare the x-coordinates of the two vertices. - Compare the constant terms in the two vertex-form equations.

1. Complete the square for \(g\): \(g(x)=x^2-4x+4-4+7\). 2. Simplify to obtain vertex form: \(g(x)=(x-2)^2+3\). 3. Therefore, the vertex of \(g\) is \((2, 3)\). 4. From the equation for \(h\), its vertex is \((2, 1)\). 5. Both vertices have x-coordinate \(2\), so both parabolas have the axis of symmetry \(x=2\). 6. Because \(g(x)=h(x)+2\), the graph of \(g\) is \(2\) units above the graph of \(h\) for every x-value. The vertical distance between the vertices is \(3-1=2\) units.

The vertex of \(g\) is \((2, 3)\). Both parabolas have the axis of symmetry \(x=2\). The graph of \(g\) is \(2\) units above the graph of \(h\), and the vertical distance between their vertices is \(2\) units.

- Isolate the squared expression first. - For a trinomial, add the value needed to create a perfect square. - When a positive square is isolated, remember both square roots. - Consider whether a real number can have a negative square.

1. For part a, \(x^2 = 25\), so \(x = \pm 5\). 2. For part b, write \(x^2 - 2.4x = -0.8\). Add \(1.44\) to both sides: \((x - 1.2)^2 = 0.64\). Then \(x - 1.2 = \pm 0.8\), so \(x = 2\) or \(x = 0.4\). 3. For part c, write \(x^2 + 4x = 5\). Add \(4\) to both sides: \((x + 2)^2 = 9\). Then \(x + 2 = \pm 3\), so \(x = 1\) or \(x = -5\). 4. For part d, \(x^2 = -8\). No real number has a negative square, so there is no real solution.

a) \(x \in \{-5, 5\}\) b) \(x \in \{0.4, 2\}\) c) \(x \in \{-5, 1\}\) d) \(\varnothing\)

- Recall that the square of a real number is never negative. - For part c, move the constant term and complete the square. - Use both square roots after isolating a square.

1. For part a, \(x^2 = \frac{1}{4}\), so \(x = \pm\frac{1}{2}\). 2. For part b, the square of any real expression is greater than or equal to zero. Therefore, \((x + 5)^2\) cannot equal \(-4\). 3. For part c, write \(x^2 - 2x = 3\). Add \(1\) to both sides: \((x - 1)^2 = 4\). Then \(x - 1 = \pm 2\), so \(x = 3\) or \(x = -1\).

a) \(x \in \left\{-\frac{1}{2}, \frac{1}{2}\right\}\) b) There is no real solution because the square of a real expression cannot be negative. c) \(x \in \{-1, 3\}\)

- Look for perfect-square trinomials. - What does vertex form look like when the vertex is on the x-axis? - What y-coordinate does every point on the x-axis have? - Complete the square when an expression is not already a perfect square.

1. For \(f(x)=x^2-14x+49\), factor the perfect-square trinomial: \(f(x)=(x-7)^2\). The vertex is \((7, 0)\), so it lies on the x-axis. 2. For \(g(x)=x^2+5x+6.25\), factor the perfect-square trinomial: \(g(x)=(x+2.5)^2\). The vertex is \((-2.5, 0)\), so it lies on the x-axis. 3. For \(h(x)=x^2-2x-1\), complete the square: \(h(x)=x^2-2x+1-2=(x-1)^2-2\). The vertex is \((1, -2)\), so it does not lie on the x-axis. 4. In general, a translation of \(y=x^2\) with vertex \((r, 0)\) has equation \(y=(x-r)^2\). Therefore, Maya's claim is correct.

Maya's claim is correct. a) Yes; the vertex is \((7, 0)\). b) Yes; the vertex is \((-2.5, 0)\). c) No; the vertex is \((1, -2)\).

- What constant makes the trinomial a perfect square? - A parabola whose vertex is on the x-axis has what minimum value? - How is the coefficient \(12\) related to the number inside the binomial?

1. If the vertex lies on the x-axis, the quadratic must be a perfect square with minimum value \(0\). 2. Compare \(x^2+12x+q\) with \((x+r)^2=x^2+2rx+r^2\). 3. Since \(2r=12\), \(r=6\). Therefore, \(q=r^2=36\). 4. The vertex form is \(f(x)=(x+6)^2\), with vertex \((-6, 0)\).

\(q=36\), and the vertex form is \(f(x)=(x+6)^2\).

- Isolate \(x^2\) first. - Move the two function expressions to one side and complete the square. - Equal outputs correspond to graph intersections.

1. Subtract \(4\) from both sides: \(x^2 = 2x - 4\). 2. Equal outputs would require \(x^2 - 2x + 4 = 0\). 3. Complete the square: \(x^2 - 2x + 4 = (x - 1)^2 + 3\). 4. Since \((x - 1)^2 + 3 > 0\) for every real \(x\), the functions never have the same output. 5. Therefore, the original equation has no real solutions, and the graphs have no intersection points.

a) \(x^2 = 2x - 4\) b) The functions never have the same output. c) No real solutions; the graphs have no intersection points.

- What form must the quadratic have when it has exactly one distinct real zero? - Which two real numbers have square \(64\)? - Compare the linear term with the expanded form of \((x-h)^2\).

1. A monic quadratic that touches the x-axis at exactly one point is a perfect square: \(g(x)=(x-h)^2=x^2-2hx+h^2\). 2. Compare constant terms: \(h^2=64\), so \(h=8\) or \(h=-8\). 3. Since \(b=-2h\), the corresponding values are \(b=-16\) and \(b=16\). 4. For the positive value \(b=16\), \(h=-8\), so \(g(x)=(x+8)^2\) and the vertex is \((-8, 0)\).

a) \(b=-16\) or \(b=16\) b) \((-8, 0)\)

- Complete the square by using half the coefficient of \(x\). - How does changing only the constant term affect the graph vertically? - Does the x-coordinate of the vertex change when the same constant is added to every output?

1. To place the vertex on the x-axis, write the function as a perfect square: \(h(x)=(x-r)^2=x^2-2rx+r^2\). 2. Compare linear coefficients: \(-2r=-9\), so \(r=4.5\). 3. Therefore, \(c=r^2=4.5^2=20.25\). 4. Increasing \(c\) by \(1.5\) adds \(1.5\) to every output: \(h_{\text{new}}(x)=(x-4.5)^2+1.5\). 5. The graph shifts \(1.5\) units upward, so the vertex moves from \((4.5, 0)\) to \((4.5, 1.5)\).

a) \(c=20.25\) b) The vertex shifts \(1.5\) units upward, from \((4.5, 0)\) to \((4.5, 1.5)\).

- What should you factor from the variable terms before completing the square? - Track the negative factor carefully when simplifying. - What do a negative leading coefficient and an absolute value between \(0\) and \(1\) do to a parabola?

1. Factor \(-0.5\) from the variable terms: \(g(x) = -0.5(x^2 - 8x) - 2\). 2. Complete the square inside the parentheses: \(g(x) = -0.5(x^2 - 8x + 16 - 16) - 2\). 3. Rewrite and simplify: \(g(x) = -0.5((x - 4)^2 - 16) - 2 = -0.5(x - 4)^2 + 8 - 2 = -0.5(x - 4)^2 + 6\). 4. The negative sign reflects the parabola across the \(x\)-axis, so it opens downward. Because \(|-0.5| < 1\), the graph is vertically compressed and therefore wider than \(y = x^2\).

The vertex form is \(g(x) = -0.5(x - 4)^2 + 6\). The graph opens downward and is wider than \(y = x^2\).

- To complete the square, add and subtract the square of half the coefficient of \(x\). - For the quadratic formula, identify \(a\), \(b\), and \(c\). - In vertex form, the vertex coordinates can be read from the expression.

1. Complete the square: \(x^2 + 8x + 12 = (x + 4)^2 - 4\). The equation becomes \((x + 4)^2 = 4\), so \(x + 4 = \pm 2\). Thus, \(x = -2\) or \(x = -6\). 2. Use the quadratic formula with \(a = 1\), \(b = 8\), and \(c = 12\): \(x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} = \frac{-8 \pm 4}{2}\). Thus, \(x = -2\) or \(x = -6\). 3. Both methods give the same solutions. Completing the square also gives the vertex form \(f(x) = (x + 4)^2 - 4\), so the vertex \((-4, -4)\) can be identified directly.

1. \(x \in \{-6, -2\}\) 2. \(x \in \{-6, -2\}\) Comparison: Completing the square produces vertex form, which shows the vertex \((-4, -4)\) directly.

- What factor should you take out of the variable terms first? - Track the outside factor when you add and subtract the value that completes the square. - Which perfect-square trinomial appears inside the parentheses?

1. Factor \(-2\) from the variable terms: \(g(x) = -2(x^2 - 6x) - 10\). 2. Complete the square inside the parentheses: \(g(x) = -2(x^2 - 6x + 9 - 9) - 10\). 3. Rewrite and simplify: \(g(x) = -2((x - 3)^2 - 9) - 10 = -2(x - 3)^2 + 18 - 10 = -2(x - 3)^2 + 8\).

\(g(x) = -2(x - 3)^2 + 8\)

- Square half the coefficient of \(x\). - Compare that square with the constant term in each equation. - A perfect-square trinomial has the form \((x - h)^2\).

1. In equation (1), half the coefficient of \(x\) is \(-5\), and \((-5)^2 = 25\). Therefore, \(x^2 - 10x + 25 = (x - 5)^2\). 2. Solve \((x - 5)^2 = 0\): \(x = 5\). 3. For equation (2), a perfect-square trinomial with middle term \(-10x\) must have constant term \(\left(\frac{-10}{2}\right)^2 = 25\). 4. Since the constant term is \(24\), it must be increased by \(1\) to make the expression a perfect-square trinomial.

a) \(x = 5\) b) The constant term must be \(25\), so \(24\) must be increased by \(1\).

- What number completes the square for \(x^2 - 4x\)? - Remember to add and subtract the same value. - Can you find the zeros directly from vertex form? - How can substitution check each zero?

1. Complete the square: \(f(x) = x^2 - 4x + 4 - 4 - 5 = (x - 2)^2 - 9\). The vertex is \((2, -9)\). 2. Set the vertex form equal to zero: \((x - 2)^2 - 9 = 0\), so \((x - 2)^2 = 9\). Then \(x - 2 = \pm 3\), giving \(x = -1\) or \(x = 5\).

a) \(f(x) = (x - 2)^2 - 9\); vertex \((2, -9)\) b) \(x = -1\) and \(x = 5\)

- What factor should you take out of the variable terms before completing the square? - Remember to distribute the outside factor to both terms when simplifying. - What signs do the coordinates of a point in Quadrant I have?

1. Factor \(-2\) from the variable terms: \(g(x) = -2(x^2 - 4x) - 5\). 2. Complete the square: \(g(x) = -2(x^2 - 4x + 4 - 4) - 5\). 3. Rewrite and simplify: \(g(x) = -2((x - 2)^2 - 4) - 5 = -2(x - 2)^2 + 8 - 5 = -2(x - 2)^2 + 3\). 4. The vertex is \((2, 3)\). Since both coordinates are positive, the vertex lies in Quadrant I.

The vertex form is \(g(x) = -2(x - 2)^2 + 3\). The vertex is \((2, 3)\), which lies in Quadrant I.

- Half of an odd coefficient will be a fraction. - Square half the coefficient of \(x\) and add that value to both sides. - Keep the work in exact fractional form.

1. For equation 1, write \(x^2 - 7x = -10\). Add \(\frac{49}{4}\) to both sides: \(\left(x - \frac{7}{2}\right)^2 = \frac{9}{4}\). 2. Take both square roots: \(x - \frac{7}{2} = \pm\frac{3}{2}\). Thus, \(x = 5\) or \(x = 2\). 3. For equation 2, add \(\frac{81}{4}\) to both sides: \(\left(x + \frac{9}{2}\right)^2 = \frac{121}{4}\). 4. Take both square roots: \(x + \frac{9}{2} = \pm\frac{11}{2}\). Thus, \(x = 1\) or \(x = -10\).

1) \(x \in \{2, 5\}\) 2) \(x \in \{-10, 1\}\)

- Isolate the terms containing \(x\). - Add the square of half the coefficient of \(x\) to both sides. - Use both square roots and simplify exactly.

1. For equation 1, write \(x^2 - 16x = -63\). Add \(64\) to both sides: \((x - 8)^2 = 1\). 2. Take both square roots: \(x - 8 = \pm 1\). Thus, \(x = 9\) or \(x = 7\). 3. For equation 2, add \(\frac{49}{4}\) to both sides: \(\left(x + \frac{7}{2}\right)^2 = \frac{121}{4}\). 4. Take both square roots: \(x + \frac{7}{2} = \pm\frac{11}{2}\). Thus, \(x = 2\) or \(x = -9\).

1) \(x \in \{7, 9\}\) 2) \(x \in \{-9, 2\}\)

- Make the coefficient of \(x^2\) equal to \(1\) before completing the square. - Add the square of half the coefficient of \(x\) to both sides. - Use exact fractions in the solution set.

1. For equation 1, divide by \(3\): \(x^2 - 4x = 12\). Add \(4\) to both sides: \((x - 2)^2 = 16\). 2. Take both square roots: \(x - 2 = \pm 4\). Thus, \(x = 6\) or \(x = -2\). 3. For equation 2, add \(\frac{9}{4}\) to both sides: \(\left(x - \frac{3}{2}\right)^2 = 4\). 4. Take both square roots: \(x - \frac{3}{2} = \pm 2\). Thus, \(x = \frac{7}{2}\) or \(x = -\frac{1}{2}\).

1) \(x \in \{-2, 6\}\) 2) \(x \in \left\{-\frac{1}{2}, \frac{7}{2}\right\}\)

- Use exact fractions rather than decimals. - Make the coefficient of \(x^2\) equal to \(1\) before completing the square. - Add the same completing-square value to both sides.

1. For part a, write \(x^2 - x = \frac{3}{4}\). Add \(\frac{1}{4}\) to both sides: \(\left(x - \frac{1}{2}\right)^2 = 1\). 2. Take both square roots: \(x - \frac{1}{2} = \pm 1\). Thus, \(x = \frac{3}{2}\) or \(x = -\frac{1}{2}\). 3. For part b, divide by \(3\): \(x^2 + 4x - 5 = 0\). Write \(x^2 + 4x = 5\). 4. Add \(4\) to both sides: \((x + 2)^2 = 9\). 5. Take both square roots: \(x + 2 = \pm 3\). Thus, \(x = 1\) or \(x = -5\).

a) \(x \in \left\{-\frac{1}{2}, \frac{3}{2}\right\}\) b) \(x \in \{-5, 1\}\)

- Calculate the discriminant from standard form. - Factor out \(0.5\) from the quadratic and linear terms before completing the square. - Read the minimum value from vertex form.

1. The discriminant is \(D = (-3)^2 - 4 \cdot 0.5 \cdot 5.5 = 9 - 11 = -2\). Since \(D < 0\), the graph has no x-intercepts. 2. Complete the square: \(f(x) = 0.5(x^2 - 6x) + 5.5 = 0.5\left((x - 3)^2 - 9\right) + 5.5\). 3. Thus, \(f(x) = 0.5(x - 3)^2 + 1\). 4. Since \((x - 3)^2 \ge 0\), it follows that \(0.5(x - 3)^2 + 1 \ge 1\). 5. The minimum value is \(1\), so the range is \([1, \infty)\).

a) \(D = -2\), so there are no x-intercepts. b) \(f(x) = 0.5(x - 3)^2 + 1\), so \(f(x) \ge 1\). c) \([1, \infty)\)

- Factor out the leading coefficient before completing the square. - Set the vertex form equal to zero and isolate the squared expression. - Use the zeros to write the factors.

1. Factor \(1.5\) from the quadratic and linear terms: \(f(x) = 1.5(x^2 - 4x) + 4.5\). 2. Complete the square: \(f(x) = 1.5((x - 2)^2 - 4) + 4.5 = 1.5(x - 2)^2 - 1.5\). 3. Set the vertex form equal to zero: \(1.5(x - 2)^2 - 1.5 = 0\). Then \((x - 2)^2 = 1\), so \(x - 2 = \pm 1\). The zeros are \(x = 1\) and \(x = 3\). 4. The factored form is \(f(x) = 1.5(x - 1)(x - 3)\). 5. The graph confirms vertex \((2, -1.5)\) and x-intercepts \((1, 0)\) and \((3, 0)\).

a) \(f(x) = 1.5(x - 2)^2 - 1.5\) b) \(x = 1\) and \(x = 3\) c) \(f(x) = 1.5(x - 1)(x - 3)\) d) The graph confirms vertex \((2, -1.5)\) and x-intercepts \((1, 0)\) and \((3, 0)\).

- How can you factor the coefficient of \(x^2\) from the variable terms? - What does vertex form tell you about a minimum or maximum? - Use the function's range to analyze part c). - What do solutions to \(h(x) = 10\) represent on the graph?

1. Factor the leading coefficient from the variable terms and complete the square: \(h(x) = 2(x^2 - 6x) + 10 = 2(x^2 - 6x + 9 - 9) + 10 = 2(x - 3)^2 - 8\). 2. The vertex is \((3, -8)\). Because the leading coefficient is positive, the parabola opens upward, so the minimum value is \(-8\) at \(x = 3\). 3. The range is \(h(x) \geq -8\). Since \(-10 < -8\), the equation \(h(x) = -10\) has no real solution. 4. Solve \(h(x) = 10\): \(2x^2 - 12x + 10 = 10\), so \(2x^2 - 12x = 0\). Factoring gives \(2x(x - 6) = 0\), so \(x = 0\) or \(x = 6\). The graph intersects the horizontal line \(y = 10\) at \((0, 10)\) and \((6, 10)\).

a) \(h(x) = 2(x - 3)^2 - 8\) b) The minimum value is \(-8\) at \(x = 3\). c) The function never takes a value below \(-8\), so it cannot equal \(-10\). d) \(x = 0\) and \(x = 6\); the intersections with \(y = 10\) are \((0, 10)\) and \((6, 10)\).

- What must the vertex's \(y\)-coordinate be if it lies on the \(x\)-axis? - Treat \(p\) like a number while completing the square. - Which solution satisfies the condition \(p > 0\)?

1. Complete the square with the parameter: \(h_p(x) = x^2 - px + \frac{p^2}{4} - \frac{p^2}{4} + 9 = \left(x - \frac{p}{2}\right)^2 + 9 - \frac{p^2}{4}\). 2. The vertex lies on the \(x\)-axis when its \(y\)-coordinate is \(0\): \(9 - \frac{p^2}{4} = 0\). 3. Then \(p^2 = 36\), so \(p = 6\) or \(p = -6\). 4. Because \(p > 0\), the required value is \(p = 6\).

\(p = 6\)

- Add and subtract the square of half the coefficient of \(x\). - Identify the vertex from the resulting vertex form. - Rewrite the claimed inequality in terms of the vertex’s y-coordinate.

1. Complete the square: \(f(x) = x^2 + px + \left(\frac{p}{2}\right)^2 - \left(\frac{p}{2}\right)^2 + q\). 2. Therefore, \(f(x) = \left(x + \frac{p}{2}\right)^2 + q - \frac{p^2}{4}\). 3. The vertex is \(\left(-\frac{p}{2}, q - \frac{p^2}{4}\right)\). 4. The condition \(q < \left(\frac{p}{2}\right)^2\) is equivalent to \(q - \frac{p^2}{4} < 0\), so the vertex lies below the x-axis. 5. Since the parabola opens upward, a vertex below the x-axis implies that the graph crosses the x-axis twice. The claim is true.

a) The vertex is \(\left(-\frac{p}{2}, q - \frac{p^2}{4}\right)\). b) The claim is true because the condition places the vertex below the x-axis while the parabola opens upward.

- View \(4a^2x^2\) as \((2ax)^2\). - Match the left side to \(A^2 + 2AB + B^2\). - After forming a square, take both square roots and isolate \(x\).

1. Multiply by \(4a\): \(4a^2x^2 + 4abx + 4ac = 0\). 2. Subtract \(4ac\): \(4a^2x^2 + 4abx = -4ac\). 3. Add \(b^2\): \(4a^2x^2 + 4abx + b^2 = b^2 - 4ac\). 4. The left side is a perfect square: \((2ax + b)^2 = b^2 - 4ac\). 5. Take both square roots: \(2ax + b = \pm\sqrt{b^2 - 4ac}\). 6. Subtract \(b\) and divide by \(2a\): \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). 7. The perfect-square form is useful because it places the variable in one squared expression that can be isolated by taking square roots.

a) \(4a^2x^2 + 4abx = -4ac\) b) \((2ax + b)^2 = b^2 - 4ac\) c) \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\); the perfect-square form allows the equation to be solved by taking square roots.