- Focus on the coefficient of the linear term. - A perfect square \((x+m)^2\) has middle term \(2mx\). - Once you identify \(m\), compare the constant term with \(m^2\).

1. Half of the coefficient of \(x\) is \(5\). 2. Squaring that value gives \(25\), so \(x^2+10x+25=(x+5)^2\).

Add \(25\); \(x^2+10x+25=(x+5)^2\).

- What terms appear if you multiply the first \(x\) by each term in the second parentheses? - Compare those with the products made by \(-4\). Which terms combine?

1. Multiply \((x-4)(x-4)=x^2-4x-4x+16\). 2. Combining like terms gives \(x^2-8x+16\); each cross-product contains \(-4\), so their sum is negative.

\(x^2-8x+16\). The middle term is \(-8x\) because the two cross-products are both \(-4x\).

- How can you compare the two expressions without trying individual values of \(x\)? - Try multiplying before combining like terms. Which products involve one \(x\)?

1. Expand the right side: \((x+6)^2=x^2+12x+36\). 2. Match the coefficients of \(x\): \(b=12\).

\(b=12\)

- Expand the student's right side to check it. - What extra constant appears when you make a complete square? - How can you balance that extra constant without changing the left side?

1. Expanding shows \((x+3)^2=x^2+6x+9\), so the student's equation adds an extra \(9\). 2. Subtract \(9\) to keep the expression unchanged: \(x^2+6x=(x+3)^2-9\).

No. The correct identity is \(x^2+6x=(x+3)^2-9\).

- Move the constant term before changing the left side. - Half the coefficient of \(x\), then square that value. - The two square roots lead to two linear equations.

1. Move the constant: \(x^2+2x=3\). 2. Add \(1\) to both sides: \(x^2+2x+1=4\). 3. Rewrite: \((x+1)^2=4\). 4. Take square roots: \(x+1=\pm2\). 5. Therefore, \(x=1\) or \(x=-3\).

\((x+1)^2=4\); \(x\in\{-3,1\}\)

- Isolate the quadratic and linear terms first. - The number added is the square of half of \(-4\). - Keep the radical exact after taking square roots.

1. Move the constant: \(x^2-4x=-1\). 2. Add \(4\): \(x^2-4x+4=3\). 3. Rewrite: \((x-2)^2=3\). 4. Take both square roots and solve: \(x=2\pm\sqrt3\).

\((x-2)^2=3\); \(x=2-\sqrt3\) or \(x=2+\sqrt3\)

- Use half of \(10\) to determine the square-completion value. - Add the same amount to both sides of the equation. - Simplify \(\sqrt8\) before writing the final roots.

1. Write \(x^2+10x=-17\). 2. Add \(25\) to both sides: \(x^2+10x+25=8\). 3. Rewrite: \((x+5)^2=8\). 4. Take square roots: \(x+5=\pm2\sqrt2\). 5. Thus, \(x=-5\pm2\sqrt2\).

\((x+5)^2=8\); \(x=-5-2\sqrt2\) or \(x=-5+2\sqrt2\)

- 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?

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 \((6,0)\). b) Since \(6.25=2.5^2\), \(x^2+5x+6.25=x^2+2\cdot2.5x+2.5^2=(x+2.5)^2\). Thus, \(g(x)=(x+2.5)^2\), and the vertex is \((-2.5,0)\). 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 \((1, 4)\)

- Connect the requirement on the vertex with a useful form of a monic quadratic. - Use the linear coefficient to determine the horizontal location of the vertex. - Check that the completed expression has minimum value \(0\).

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

- Square half the coefficient of \(x\). - Add and subtract that same number so the function is unchanged. - The sign of the squared-term coefficient determines whether the vertex is a minimum or maximum.

1. Add and subtract \(25\): \(f(x)=x^2+10x+25-25+7\). 2. Rewrite: \(f(x)=(x+5)^2-18\). 3. The vertex is \((-5,-18)\). Because the leading coefficient is positive, it is a minimum.

Completion line: \(f(x)=x^2+10x+25-25+7\) Vertex form: \(f(x)=(x+5)^2-18\) Vertex: \((-5,-18)\), a minimum.

- Focus first on the terms \(x^2+4x\). - The number used to complete the square comes from half the coefficient of \(x\), then squaring. - Keep the expression equivalent by compensating for whatever constant you insert.

1. Half of the coefficient of \(x\) is \(2\), and \(2^2=4\), so add and subtract \(4\): \(x^2+4x+7=x^2+4x+4-4+7\). 2. Rewrite the perfect-square trinomial: \(f(x)=(x+2)^2+3\). 3. From vertex form, the vertex is \((-2,3)\).

Completing-square step: \(x^2+4x+7=x^2+4x+4-4+7\) Vertex form: \(f(x)=(x+2)^2+3\) Vertex: \((-2,3)\)

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

a) In equation (1), half the coefficient of \(x\) is \(-5\), and \((-5)^2=25\). Therefore, \(x^2-10x+25=(x-5)^2\). Solving \((x-5)^2=0\) gives \(x=5\). b) A perfect-square trinomial with middle term \(-10x\) must have constant term \(\left(\frac{-10}{2}\right)^2=25\). Since equation (2) has constant term \(24\), it must be increased by \(1\).

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

- Move the constant term before completing the square. - The square of half of \(2\) is the value to add. - Keep both square-root branches in exact form.

1. Write \(x^2+2x=4\). 2. Add \(1\): \((x+1)^2=5\). 3. Take square roots: \(x+1=\pm\sqrt5\). 4. Thus, \(x=-1\pm\sqrt5\).

\((x+1)^2=5\); \(x=-1-\sqrt5\) or \(x=-1+\sqrt5\)

- Isolate the quadratic and linear terms first. - Complete the square using half of the linear coefficient. - Simplify \(\sqrt8\) after isolating the square.

1. Write \(x^2+4x=4\). 2. Add \(4\): \((x+2)^2=8\). 3. Take square roots: \(x+2=\pm2\sqrt2\). 4. Therefore, \(x=-2\pm2\sqrt2\).

\((x+2)^2=8\); \(x=-2-2\sqrt2\) or \(x=-2+2\sqrt2\)

- Put each equation in the form \(x^2+bx=c\). - Add the square of half of \(b\) to both sides. - Do not skip the completed-square equation when reporting your work.

1. Write \(x^2+10x=24\), add \(25\), and obtain \((x+5)^2=49\). Thus, \(x=-12\) or \(x=2\). 2. Add \(64\) to both sides to obtain \((x-8)^2=49\). Thus, \(x=1\) or \(x=15\).

1) \((x+5)^2=49\); \(x\in\{-12,2\}\) 2) \((x-8)^2=49\); \(x\in\{1,15\}\)

- Move the constant term before completing each square. - Use the square of half of the linear coefficient. - Report the square you formed before using its two roots.

1. a) Write \(x^2+4x=21\), add \(4\), and get \((x+2)^2=25\). Thus, \(x=-7\) or \(x=3\). 2. b) Write \(x^2-6x=-5\), add \(9\), and get \((x-3)^2=4\). Thus, \(x=1\) or \(x=5\).

a) \((x+2)^2=25\); \(x\in\{-7,3\}\) b) \((x-3)^2=4\); \(x\in\{1,5\}\)

- Use the two labeled side lengths that bound the shaded corner. - Think of \(6x\) as the combined area of the two rectangular strips. - Compare the completed figure with the area formula for one large square.

1. The shaded corner has side lengths \(3\) and \(3\), so its area is \(9\). 2. The other three regions have total area \(x^2+3x+3x=x^2+6x\). 3. Adding the corner gives \(x^2+6x+9\), which is the area of a square with side length \(x+3\). Therefore, \(x^2+6x+9=(x+3)^2\).

a) \(9\) b) The added \(3\times3\) corner completes a square of side \(x+3\), so \(x^2+6x+9=(x+3)^2\).

- After taking out \(-1\), which signs belong inside the parentheses? - What number belongs in the squared binomial so that its middle term matches? - The extra constant is also affected by the negative sign outside.

1. Take out the negative sign: \(-(x^2-8x)+1\). 2. Inside the parentheses, \(x^2-8x=(x-4)^2-16\). 3. Therefore \(-x^2+8x+1=-((x-4)^2-16)+1=-(x-4)^2+17\).

First step: \(-(x^2-8x)+1\). Final form: \(-(x-4)^2+17\).

- Expand the student's proposed expression to check all its coefficients. - Account for the number multiplied outside the square when adjusting the constant. - What constant must remain after expansion?

1. Expand the proposed form: \(2(x+3)^2-4=2x^2+12x+14\), which has the wrong constant. 2. Since \(2(x+3)^2=2x^2+12x+18\), subtract \(13\) to get \(2x^2+12x+5=2(x+3)^2-13\).

The constant is too large. The correct form is \(2(x+3)^2-13\).

- Look at the lowest point of the short curve. What height does it reach? - How can you change \(x^2+4x\) into a square without changing its value? - What does the number outside the square do to a graph's height?

1. The vertex is \((-2, -1)\). 2. Add and subtract \(4\): \(x^2+4x+c=(x^2+4x+4)+c-4=(x+2)^2+c-4\). 3. The vertex's y-coordinate is \(c-4=-1\), so \(c=3\) and \(f(x)=(x+2)^2-1\).

\(x^2+4x+c=(x^2+4x+4)+c-4=(x+2)^2+c-4\). The vertex is \((-2, -1)\), so \(c-4=-1\) and \(c=3\).

- Make the coefficient of \(x^2\) equal to \(1\) before completing the square. - Add the square of half the linear coefficient to both sides. - Preserve both signs when taking the square root.

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

Completed-square equation: \((x-2)^2=\frac{25}{4}\) Solutions: \(x=-\frac{1}{2}\) or \(x=\frac{9}{2}\)

- Normalize the quadratic coefficient before completing the square. - Keep fractional constants exact while adding the completion value. - Use both signs after taking square roots.

1. Move \(29\) and divide by \(4\): \(x^2-6x=-\frac{29}{4}\). 2. Add \(9\): \(x^2-6x+9=\frac{7}{4}\). 3. Rewrite: \((x-3)^2=\frac{7}{4}\). 4. Take both square roots: \(x-3=\pm\frac{\sqrt7}{2}\). 5. Therefore, \(x=3\pm\frac{\sqrt7}{2}\).

\((x-3)^2=\frac{7}{4}\); \(x=3-\frac{\sqrt7}{2}\) or \(x=3+\frac{\sqrt7}{2}\)

- Factor the leading coefficient only from the terms containing \(x\). - Complete the square inside the parentheses before distributing the outside factor. - Read horizontal and vertical shifts from the final vertex form.

1. Factor \(2\) from the variable terms: \(f(x)=2(x^2-6x)+10\). 2. Complete the square: \(f(x)=2((x-3)^2-9)+10\). 3. Simplify: \(f(x)=2(x-3)^2-8\). 4. The vertex is \((3,-8)\). 5. Relative to \(y=x^2\), the graph is vertically stretched by \(2\), shifted \(3\) units right, and shifted \(8\) units down.

Completion step: \(f(x)=2((x-3)^2-9)+10\) Vertex form: \(f(x)=2(x-3)^2-8\) Vertex: \((3,-8)\) Transformations: vertical stretch by \(2\), right \(3\), down \(8\).

- Add and subtract the same square-completion value within \(g\). - Compare the vertex x-coordinates after rewriting \(g\). - Subtract the two vertex-form expressions to compare their heights.

1. Complete the square: \(g(x)=x^2-4x+4-4+7=(x-2)^2+3\). 2. The vertex of \(g\) is \((2,3)\); the vertex of \(h\) is \((2,1)\). 3. Both have axis \(x=2\). 4. Since \(g(x)=h(x)+2\), \(g\) is \(2\) units above \(h\) everywhere and the vertices are \(2\) units apart vertically.

Completion: \(g(x)=x^2-4x+4-4+7=(x-2)^2+3\) Both axes are \(x=2\). The graph of \(g\) is \(2\) units above \(h\), and the vertices are \(2\) units apart vertically.

- Decide first whether the square is already isolated or whether a linear term requires completion. - For b) and c), square half the coefficient of \(x\). - A real square cannot equal a negative number.

1. a) \(x^2=25\), so \(x=\pm5\). 2. b) Write \(x^2-2.4x=-0.8\) and add \(1.44\): \((x-1.2)^2=0.64\). Thus, \(x=0.4\) or \(x=2\). 3. c) Write \(x^2+4x=5\) and add \(4\): \((x+2)^2=9\). Thus, \(x=-5\) or \(x=1\). 4. d) \(x^2=-8\), which has no real solution.

a) \(x^2=25\); \(x\in\{-5,5\}\) b) \((x-1.2)^2=0.64\); \(x\in\{0.4,2\}\) c) \((x+2)^2=9\); \(x\in\{-5,1\}\) d) \(x^2=-8\); \(\varnothing\)

- Use the sign restriction on real squares in part b). - For part c), isolate the quadratic and linear terms before adding the completion value. - Keep both roots after the completed square is isolated.

1. a) \(x^2=\frac14\), so \(x=\pm\frac12\). 2. b) The square of any real expression is nonnegative, so it cannot equal \(-4\). 3. c) Write \(x^2-2x=3\), add \(1\), and obtain \((x-1)^2=4\). Thus, \(x=-1\) or \(x=3\).

a) \(x\in\left\{-\frac12,\frac12\right\}\) b) No real solution because a real square cannot be negative. c) \((x-1)^2=4\); \(x\in\{-1,3\}\)

- Rewrite each example in vertex form before judging the claim. - In the general form, identify which parameter is the vertex's y-coordinate. - Ask what setting that coordinate to zero does to the equation.

1. a) \(f(x)=(x-7)^2\), so the vertex is \((7,0)\). 2. b) \(g(x)=(x+2.5)^2\), so the vertex is \((-2.5,0)\). 3. c) Completing the square gives \(h(x)=(x-1)^2-2\), so the vertex is \((1,-2)\). 4. A general translation has vertex \((h,k)\). Its vertex lies on the x-axis exactly when \(k=0\). 5. When \(k=0\), the equation becomes \(y=(x-h)^2\), the square of a binomial. Therefore, Maya's claim is true for every translation of \(y=x^2\).

a) \((x-7)^2\); vertex \((7,0)\) b) \((x+2.5)^2\); vertex \((-2.5,0)\) c) \((x-1)^2-2\); vertex \((1,-2)\) General proof: \(y=(x-h)^2+k\) has vertex \((h,k)\). A vertex on the x-axis requires \(k=0\), giving \(y=(x-h)^2\). Therefore, Maya's claim is true.

- Add and subtract the square of half of \(12\). - Read the vertex's y-coordinate from the completed-square form. - A point on the x-axis has y-coordinate \(0\).

1. Complete the square: \(f(x)=x^2+12x+36-36+q=(x+6)^2+(q-36)\). 2. The vertex is \((-6,q-36)\). 3. For the vertex to lie on the x-axis, \(q-36=0\), so \(q=36\). 4. Then \(f(x)=(x+6)^2\).

General completed-square form: \(f(x)=(x+6)^2+(q-36)\) \(q=36\), so \(f(x)=(x+6)^2\).

- Move the two output expressions to one side before completing the square. - A squared expression is always nonnegative over the reals. - Equal function outputs correspond to intersection points.

1. a) Subtract \(4\): \(x^2=2x-4\). 2. Equal outputs require \(x^2-2x+4=0\). Complete the square: \(x^2-2x+4=(x-1)^2+3\). 3. Since \((x-1)^2+3>0\) for every real \(x\), equal outputs are impossible. 4. Therefore, the original equation has no real solutions and the graphs do not intersect.

a) \(x^2=2x-4\) b) \(f(x)-g(x)=(x-1)^2+3>0\), so the functions never have the same output. c) No real solutions; the graphs have no intersection points.

- Treat \(b\) as a fixed number while completing the square in \(x\). - Which part of the completed-square form tells you the lowest height of this upward-opening graph? - Imagine raising or lowering that graph. When would it have just one point on the x-axis?

1. Complete the square: \(g(x)=\left(x+\frac{b}{2}\right)^2+64-\frac{b^2}{4}\). 2. The graph touches the x-axis once when the vertex y-coordinate is \(0\): \(64-\frac{b^2}{4}=0\). 3. Thus, \(b^2=256\), so \(b=-16\) or \(b=16\). 4. For \(b=16\), \(g(x)=(x+8)^2\), whose vertex is \((-8, 0)\).

a) \(g(x)=\left(x+\frac{b}{2}\right)^2+64-\frac{b^2}{4}\); \(b=-16\) or \(b=16\). b) For \(b=16\), the vertex is \((-8, 0)\).

- Expand \((x-m)^2\) and compare the coefficient of \(x\) with the original expression. - After matching the \(x\)-terms, which constant must be adjusted to keep the expression equal? - Compare the old and new function values at the same input. What happens to the vertex?

1. Complete the square: \(h(x)=(x-4.5)^2+c-20.25\). 2. The vertex is \((4.5, c-20.25)\), so it lies on the x-axis when \(c=20.25\). 3. Increasing \(c\) by \(1.5\) shifts every output up by \(1.5\), moving the vertex to \((4.5, 1.5)\).

a) \(h(x)=(x-4.5)^2+c-20.25\); \(c=20.25\). b) The vertex moves from \((4.5, 0)\) to \((4.5, 1.5)\), a shift of \(1.5\) units up.

- Factor the leading coefficient from the variable terms first. - Track the outside negative factor when you complete the square inside parentheses. - Read both the horizontal and vertical translations from the completed vertex form.

1. Factor \(-0.5\): \(g(x)=-0.5(x^2-8x)-2\). 2. Complete the square: \(g(x)=-0.5((x-4)^2-16)-2\). 3. Simplify: \(g(x)=-0.5(x-4)^2+6\). 4. Relative to \(y=x^2\), reflect across the x-axis, shrink vertically by a factor of \(0.5\), shift right \(4\) units and up \(6\) units.

Completion step: \(g(x)=-0.5((x-4)^2-16)-2\) Vertex form: \(g(x)=-0.5(x-4)^2+6\) Relative to \(y=x^2\), the graph is reflected across the x-axis, vertically shrunk by a factor of \(0.5\), and shifted right \(4\) units and up \(6\) units.

- For the first method, turn the quadratic and linear terms into one square before solving. - For the second method, identify all three coefficients with their signs. - Compare what geometric information is visible in the completed-square form but not in the raw formula substitution.

1. Complete the square: \(x^2+8x+12=(x+4)^2-4\), so \((x+4)^2=4\). Thus, \(x=-6\) or \(x=-2\). 2. With \(a=1\), \(b=8\), and \(c=12\), the quadratic formula gives \(x=\frac{-8\pm\sqrt{8^2-4\cdot1\cdot12}}{2\cdot1}=\frac{-8\pm4}{2}\), giving the same roots. 3. Completing the square also produces \(f(x)=(x+4)^2-4\), from which the vertex \((-4,-4)\) is visible directly.

1. Completing the square: \((x+4)^2=4\); \(x\in\{-6,-2\}\) 2. Quadratic formula: \(x=\frac{-8\pm\sqrt{8^2-4\cdot1\cdot12}}{2\cdot1}\); \(x\in\{-6,-2\}\) Comparison: Completing the square also gives vertex form \((x+4)^2-4\), which shows the vertex \((-4,-4)\) directly.

- Take the negative leading coefficient outside the variable terms first. - Complete the square inside the parentheses while keeping the outside factor in view. - Distribute the outside factor only after the square has been formed.

1. Factor \(-2\) from the variable terms: \(g(x)=-2(x^2-6x)-10\). 2. Complete the square: \(g(x)=-2((x-3)^2-9)-10\). 3. Simplify: \(g(x)=-2(x-3)^2+8\).

Completion step: \(g(x)=-2((x-3)^2-9)-10\) Vertex form: \(g(x)=-2(x-3)^2+8\)

- Match the leading term to \(a^2x^2\). - Use the middle term \(2abx\) to find \(b\). - The constant term is \(b^2\).

1. Compare with \((ax + b)^2 = a^2x^2 + 2abx + b^2\). 2. Since \(a^2 = 4\) and \(a > 0\), \(a = 2\). 3. The middle term gives \(2(2)b = 12\), so \(b = 3\). 4. Therefore, \(c = b^2 = 9\), and the expression factors as \((2x + 3)^2\).

\(c = 9\); \((2x + 3)^2\)

- Add and subtract the square of half of \(-4\). - Read the vertex only after the function is in completed-square form. - For the zeros, set the completed-square expression equal to zero and isolate the square.

1. Add and subtract \(4\): \(f(x)=x^2-4x+4-4-5=(x-2)^2-9\). 2. The vertex is \((2,-9)\). 3. Set the vertex form equal to zero: \((x-2)^2=9\), so \(x=-1\) or \(x=5\).

a) Completion: \(x^2-4x+4-4-5\); \(f(x)=(x-2)^2-9\); vertex \((2,-9)\) b) \(x=-1\) and \(x=5\)

- Factor the leading coefficient from the terms containing \(x\). - Complete the square inside the parentheses before applying the outside factor. - Determine the quadrant from the signs of the two vertex coordinates.

1. Factor \(-2\): \(g(x)=-2(x^2-4x)-5\). 2. Complete the square: \(g(x)=-2((x-2)^2-4)-5\). 3. Simplify: \(g(x)=-2(x-2)^2+3\). 4. The vertex is \((2,3)\), in Quadrant I.

Completion step: \(g(x)=-2((x-2)^2-4)-5\) Vertex form: \(g(x)=-2(x-2)^2+3\) Vertex: \((2,3)\), Quadrant I.

- Treat \(p\) as a constant while completing the square in \(x\). - Which term in vertex form gives the vertex's y-coordinate? - Use the restriction on \(p\) to choose between values obtained from a square.

1. Complete the square: \(h_p(x)=\left(x-\frac{p}{2}\right)^2+9-\frac{p^2}{4}\). 2. The vertex y-coordinate is \(9-\frac{p^2}{4}\). 3. Set it equal to zero: \(p^2=36\). 4. Since \(p>0\), \(p=6\).

Completed-square form: \(h_p(x)=\left(x-\frac{p}{2}\right)^2+9-\frac{p^2}{4}\) \(p=6\)

- When the coefficient of \(x\) is odd, half of it is a fraction. - Keep the fraction exact while squaring it and adding it to both sides. - Check the completed-square equation before taking roots.

1. Write \(x^2-7x=-10\), add \(\frac{49}{4}\), and get \(\left(x-\frac{7}{2}\right)^2=\frac{9}{4}\). Thus, \(x=2\) or \(x=5\). 2. Add \(\frac{81}{4}\): \(\left(x+\frac{9}{2}\right)^2=\frac{121}{4}\). Thus, \(x=-10\) or \(x=1\).

1) \(\left(x-\frac{7}{2}\right)^2=\frac{9}{4}\); \(x\in\{2,5\}\) 2) \(\left(x+\frac{9}{2}\right)^2=\frac{121}{4}\); \(x\in\{-10,1\}\)

- Move the constant term away from the quadratic and linear terms first. - Square half of the linear coefficient, even when that half is fractional. - Take both square roots only after a single square is isolated.

1. Write \(x^2-16x=-63\), add \(64\), and obtain \((x-8)^2=1\). Thus, \(x=7\) or \(x=9\). 2. Add \(\frac{49}{4}\): \(\left(x+\frac{7}{2}\right)^2=\frac{121}{4}\). Thus, \(x=-9\) or \(x=2\).

1) \((x-8)^2=1\); \(x\in\{7,9\}\) 2) \(\left(x+\frac{7}{2}\right)^2=\frac{121}{4}\); \(x\in\{-9,2\}\)

- Normalize the leading coefficient before completing the square when necessary. - Work with exact fractions in part 2). - Check that the expression on the left is a squared binomial before solving.

1. Divide by \(3\): \(x^2-4x=12\). Add \(4\): \((x-2)^2=16\). Thus, \(x=-2\) or \(x=6\). 2. Add \(\frac{9}{4}\): \(\left(x-\frac{3}{2}\right)^2=4\). Thus, \(x=-\frac{1}{2}\) or \(x=\frac{7}{2}\).

1) \((x-2)^2=16\); \(x\in\{-2,6\}\) 2) \(\left(x-\frac{3}{2}\right)^2=4\); \(x\in\left\{-\frac{1}{2},\frac{7}{2}\right\}\)

- Keep fractions exact instead of converting them to decimals. - Normalize part b) before adding a square-completion value. - Confirm that the left side is a perfect square before taking square roots.

1. Write \(x^2-x=\frac{3}{4}\), add \(\frac{1}{4}\), and get \(\left(x-\frac{1}{2}\right)^2=1\). Thus, \(x=-\frac{1}{2}\) or \(x=\frac{3}{2}\). 2. Divide by \(3\): \(x^2+4x-5=0\), so \(x^2+4x=5\). Add \(4\): \((x+2)^2=9\). Thus, \(x=-5\) or \(x=1\).

a) \(\left(x-\frac{1}{2}\right)^2=1\); \(x\in\left\{-\frac{1}{2},\frac{3}{2}\right\}\) b) \((x+2)^2=9\); \(x\in\{-5,1\}\)

- Before completing the square, factor the leading coefficient from the quadratic and linear terms only. - Inside the parentheses, determine the constant that turns \(x^2-6x\) into a perfect-square trinomial. - After reaching vertex form, use the fact that a real square cannot be negative.

1. Factor \(0.5\) from the quadratic and linear terms: \(f(x)=0.5(x^2-6x)+5.5\). 2. Complete the square inside the parentheses: \(f(x)=0.5((x-3)^2-9)+5.5=0.5(x-3)^2+1\). 3. Because \((x-3)^2\ge0\), \(0.5(x-3)^2+1\ge1\). Therefore \(f(x)\) can never equal \(0\), so the graph has no x-intercepts. 4. The minimum value is \(1\), so the range is \([1,\infty)\).

a) \(f(x)=0.5(x^2-6x)+5.5=0.5((x-3)^2-9)+5.5=0.5(x-3)^2+1\) b) Since \((x-3)^2\ge0\), \(f(x)\ge1\); therefore the graph has no x-intercepts. c) \([1,\infty)\)

- What perfect-square trinomial begins with \(x^2 + 4x\)? - How does the constant inside the parentheses relate to \(c\)? - What is the vertex of a quadratic written as \(a(x - h)^2\)?

1. Factor \(2\): \(f(x) = 2\left(x^2 + 4x + \frac{c}{2}\right)\). 2. The trinomial inside the parentheses must be \((x + 2)^2 = x^2 + 4x + 4\). Therefore, \(\frac{c}{2} = 4\), so \(c = 8\). 3. Then \(f(x) = 2(x + 2)^2\), whose vertex is \((-2, 0)\).

1) \(c = 8\) 2) \((-2, 0)\)

- Factor the leading coefficient before forming the square. - Use the completed-square form itself to solve for the zeros. - Once the zeros are known, include the original leading coefficient in factored form.

1. Factor \(1.5\): \(f(x)=1.5(x^2-4x)+4.5\). 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: \((x-2)^2=1\), giving zeros \(1\) and \(3\). 4. Therefore, \(f(x)=1.5(x-1)(x-3)\).

a) Completion: \(f(x)=1.5((x-2)^2-4)+4.5\); vertex form \(f(x)=1.5(x-2)^2-1.5\) b) \(x=1\) and \(x=3\) c) \(f(x)=1.5(x-1)(x-3)\)

- Track the effect of the outside coefficient on any value inserted inside the parentheses. - Preserve equality before rewriting a trinomial as a square. - Isolate the completed square before taking both square roots.

1. Inserting \(4\) inside parentheses multiplied by \(2\) increases the left side by \(8\), not \(4\), so Noah's equation is not equivalent to the original. 2. From \(2x^2+8x=3\), write \(2(x^2+4x)=3\). 3. Complete the square: \(2((x+2)^2-4)=3\), so \(2(x+2)^2=11\), or \((x+2)^2=\frac{11}{2}\). 4. Thus, \(x=-2\pm\frac{\sqrt{22}}{2}\).

Noah compensates by \(4\) even though adding \(4\) inside a quantity multiplied by \(2\) changes the left side by \(8\). Correct completed-square equation: \((x+2)^2=\frac{11}{2}\) Solutions: \(x=-2\pm\frac{\sqrt{22}}{2}\)

- Factor the leading coefficient from only the variable terms before completing the square. - Use vertex form to read the range and minimum directly. - Interpret an equation \(h(x)=c\) as intersections with a horizontal line.

1. Complete the square: \(h(x)=2(x^2-6x)+10=2((x-3)^2-9)+10=2(x-3)^2-8\). 2. The vertex is \((3,-8)\), so the minimum value is \(-8\) at \(x=3\). 3. Since the range is \(h(x)\ge-8\), the equation \(h(x)=-10\) has no real solution. 4. Solve \(h(x)=10\): \(2x^2-12x=0\), so \(2x(x-6)=0\). Thus, \(x=0\) or \(x=6\), corresponding to intersections \((0,10)\) and \((6,10)\).

a) \(h(x)=2((x-3)^2-9)+10=2(x-3)^2-8\) b) Minimum \(-8\) at \(x=3\) c) The range is \(h(x)\ge-8\), so \(h(x)=-10\) has no real solution. d) \(x=0,6\); the graph meets \(y=10\) at \((0,10)\) and \((6,10)\).

- Treat \(p\) and \(q\) as constants while completing the square in \(x\). - Compare the claim's inequality with the constant outside the square. - How does the sign of that constant affect the x-intercepts of an upward-opening parabola?

1. Complete the square: \(f(x)=\left(x+\frac{p}{2}\right)^2+q-\frac{p^2}{4}\). 2. The vertex is \(\left(-\frac{p}{2},q-\frac{p^2}{4}\right)\). 3. The stated condition is equivalent to \(q-\frac{p^2}{4}<0\), so the vertex is below the x-axis. 4. Since the parabola opens upward, it must cross the x-axis twice. The claim is true.

a) \(f(x)=\left(x+\frac{p}{2}\right)^2+q-\frac{p^2}{4}\); vertex \(\left(-\frac{p}{2},q-\frac{p^2}{4}\right)\) b) True. The vertex is below the x-axis and the parabola opens upward, so there are two x-intercepts.

- Look for the square of a binomial whose first term is \(2ax\). - What must you add to build the final constant term of that square? - Before taking a square root over the real numbers, what must be true of the other side?

1. Multiply by \(4a\): \(4a^2x^2+4abx+4ac=0\). Subtract \(4ac\): \(4a^2x^2+4abx=-4ac\). 2. Add \(b^2\): \(4a^2x^2+4abx+b^2=b^2-4ac\), so \((2ax+b)^2=b^2-4ac\). 3. If \(b^2-4ac\ge0\), take both square roots: \(2ax+b=\pm\sqrt{b^2-4ac}\). Since \(a\ne0\), \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). 4. The formula gives two distinct real roots for \(b^2-4ac>0\), one repeated root for \(b^2-4ac=0\), and no real roots for \(b^2-4ac<0\). Completing the square isolates the variable inside one square.

a) \(4a^2x^2+4abx=-4ac\) b) \((2ax+b)^2=b^2-4ac\) c) \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) for \(b^2-4ac\ge0\); if \(b^2-4ac<0\), there are no real solutions. The completed square allows the square-root step.