Completing the Square 2

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Question 1 of 5

1. Question
Solve by completing the square

$2 x^{2} - 12 x = - 8$ $2x^2-12x=-8$
- 1. $5 \pm \sqrt{2}$ $5+-sqrt2$
- 2. $- 2$ $-2$ and $3$ $3$
- 3. $1 \pm \sqrt{3}$ $1+-sqrt3$
- 4. $3 \pm \sqrt{5}$ $3+-sqrt5$
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Completing the square is done by taking the coefficient of $x$ $x$ , halving it and then squaring it. Then we add the new value to both sides of the equation.

Simplify the equation by dividing both sides by $2$ $2$

$2 x^{2} - 12 x$ $2x^2-12x$ $=$ $=$ $- 8$ $-8$

$(2 x^{2} - 12 x)$ $(2x^2-12x)$ $\div 2$ $-:2$ $=$ $=$ $- 8$ $-8$ $\div 2$ $-:2$

$x^{2} - 6 x$ $x^2-6x$ $=$ $=$ $- 4$ $-4$

Take the coefficient of the middle term, divide it by two and then square it.

$x^{2}$ $x^2$ $- 6$ $-6$ $x$ $x$ $=$ $=$ $- 4$ $-4$ Coefficient of the middle term

$- 6$ $-6$ $\div 2$ $-:2$ $=$ $=$ $- 3$ $-3$ Divide it by $2$ $2$

${(- 3)}^{2}$ $(-3)^2$ $=$ $=$ $9$ $9$ Square

This number will make a perfect square on the left side.

Add $9$ $9$ to both sides of the equation to keep the balance, then form a square of a binomial

$x^{2} - 6 x$ $x^2-6x$ $=$ $=$ $- 4$ $-4$

$x^{2} - 6 x$ $x^2-6x$ $+ 9$ $+9$ $=$ $=$ $- 4$ $-4$ $+ 9$ $+9$

${(x - 3)}^{2}$ $(x-3)^2$ $=$ $=$ $5$ $5$

Finally, take the square root of both sides and continue solving for $x$ $x$ .

${(x - 3)}^{2}$ $(x-3)^2$ $=$ $=$ $5$ $5$

$x - 3$ $x-3$ $=$ $=$ $\sqrt{5}$ $sqrt(5)$

$x - 3$ $x-3$ $+ 3$ $+3$ $=$ $=$ $\pm \sqrt{5}$ $+-sqrt(5)$ $+ 3$ $+3$ Add $3$ $3$ to both sides

$x$ $x$ $=$ $=$ $3 \pm \sqrt{5}$ $3+-sqrt5$

The roots can also be written individually

$=$ $=$ $3 + \sqrt{5}$ $3+sqrt5$

$=$ $=$ $3 - \sqrt{5}$ $3-sqrt5$

$3 \pm \sqrt{5}$ $3+-sqrt5$

Question 2 of 5

Solve by completing the square

4 x^{2} + 8 x - 7 = 0

$4x^2+8x-7=0$

1. $1 \pm \frac{\sqrt{9}}{2}$ $1+-(sqrt9)/2$
2. $- 1 \pm \frac{\sqrt{11}}{2}$ $-1+-(sqrt11)/2$
3. $2 \pm \frac{\sqrt{9}}{2}$ $2+-(sqrt9)/2$
4. $- 3 \pm \frac{\sqrt{3}}{5}$ $-3+-(sqrt3)/5$

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Completing the square is done by taking the coefficient of

x

$x$ , halving it and then squaring it. Then we add the new value to both sides of the equation.

Divide the equation by

4

$4$ to reduce the coefficient of

x^{2}

$x^2$

$4 x^{2} + 8 x - 7$ $4x^2+8x-7$	$=$ $=$	$0$ $0$
$4 x^{2} + 8 x - 7$ $4x^2+8x-7$ $\div 4$ $-:4$	$=$ $=$	$0$ $0$ $\div 4$ $-:4$

$x^{2} + 2 x - \frac{7}{4}$ $x^2+2x-7/4$	$=$ $=$	$0$ $0$

Take the coefficient of the middle term, divide it by two and then square it.

$x^{2} +$ $x^2+$ $2$ $2$ $x - \frac{7}{4}$ $x-7/4$	$=$ $=$	$0$ $0$			Coefficient of the middle term

$2$ $2$ $\div 2$ $-:2$	$=$ $=$	$1$ $1$			Divide it by $2$ $2$
		${(1)}^{2}$ $(1)^2$	$=$ $=$	$1$ $1$	Square

This number will make a perfect square on the left side.

Add and subtract $1$ $1$ to the left side of the equation to keep the balance, then form a square of a binomial

$x^{2} + 2 x - \frac{7}{4}$ $x^2+2x-7/4$	$=$ $=$	$0$ $0$

$x^{2} + 2 x$ $x^2+2x$ $+ 1 - 1$ $+1-1$ $- \frac{7}{4}$ $-7/4$	$=$ $=$	$0$ $0$

${(x + 1)}^{2} - 1 - \frac{7}{4}$ $(x+1)^2-1-7/4$	$=$ $=$	$0$ $0$
${(x + 1)}^{2} - \frac{11}{4}$ $(x+1)^2-11/4$	$=$ $=$	$0$ $0$

Move the constant to the right

${(x + 1)}^{2} - \frac{11}{4}$ $(x+1)^2-11/4$	$=$ $=$	$0$ $0$

${(x + 1)}^{2} - \frac{11}{4}$ $(x+1)^2-11/4$ $+ \frac{11}{4}$ $+11/4$	$=$ $=$	$0$ $0$ $+ \frac{11}{4}$ $+11/4$	Add $\frac{11}{4}$ $11/4$ to both sides

${(x + 1)}^{2}$ $(x+1)^2$	$=$ $=$	$\frac{11}{4}$ $11/4$

Finally, take the square root of both sides and continue solving for

x

$x$ .

${(x + 1)}^{2}$ $(x+1)^2$	$=$ $=$	$\frac{11}{4}$ $11/4$

$\sqrt{{(x + 1)}^{2}}$ $sqrt((x+1)^2)$	$=$ $=$	$\sqrt{\frac{11}{4}}$ $sqrt(11/4)$

$x + 1$ $x+1$	$=$ $=$	$\pm \frac{\sqrt{11}}{2}$ $+-(sqrt11)/2$

$x + 1$ $x+1$ $- 1$ $-1$	$=$ $=$	$\pm \frac{\sqrt{11}}{2}$ $+-(sqrt11)/2$ $- 1$ $-1$	Subtract $1$ $1$ from both sides

$x$ $x$	$=$ $=$	$- 1 \pm \frac{\sqrt{11}}{2}$ $-1+-(sqrt11)/2$

The roots can also be written individually

=

$=$

- 1 + \frac{\sqrt{11}}{2}

$-1+(sqrt11)/2$

=

$=$

- 1 - \frac{\sqrt{11}}{2}

$-1-(sqrt11)/2$

- 1 \pm \frac{\sqrt{11}}{2}

$-1+-(sqrt11)/2$

Question 3 of 5

Solve by completing the square

- 4 x^{2} + 21 x = x + 5

$-4x^2+21x=x+5$

1. $3 \pm \sqrt{5}$ $3+-sqrt5$
2. $2 \pm \sqrt{3}$ $2+-sqrt3$
3. $5 \pm \sqrt{\frac{5}{2}}$ $5+-sqrt(5/2)$
4. $\frac{5}{2} \pm \sqrt{5}$ $5/2+-sqrt5$

Question 4 of 5

Solve by completing the square

9 x^{2} + 10 x - 100 = 4 x^{2} + 15

$9x^2+10x-100=4x^2+15$

1. $- 1 \pm \sqrt{5}$ $-1+-sqrt5$
2. $- 2 \pm \sqrt{6}$ $-2+-sqrt6$
3. $1 \pm 2 \sqrt{3}$ $1+-2sqrt3$
4. $- 1 \pm 2 \sqrt{6}$ $-1+-2sqrt6$

Question 5 of 5

Find the vertex of the function:

y = 3 x^{2} - 9 x + 4

$y=3x^2-9x+4$

1. $(1 \frac{1}{2}, - 2 \frac{3}{4})$ $(1 1/2, -2 3/4)$
2. $(- 1 \frac{1}{2}, - 2 \frac{3}{4})$ $(-1 1/2, -2 3/4)$
3. $(\frac{1}{2}, 2 \frac{3}{4})$ $(1/2, 2 3/4)$
4. $(1 \frac{1}{2}, \frac{3}{4})$ $(1 1/2, 3/4)$

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Vertex Form

y = a {(x - h)}^{2} + k

$y=a(x-h)^2+k$

where

(h, k)

$(h,k)$ is the vertex

Completing the square is done by taking the coefficient of

x

$x$ , halving it and then squaring it. Then we add the new value to both sides of the equation.

To find the vertex, transform the given function into vertex form

Start by leaving

x

$x$ terms on the right side and factoring it out

$y$ $y$	$=$ $=$	$3 x^{2} - 9 x + 4$ $3x^2-9x+4$
$y$ $y$ $- 4$ $-4$	$=$ $=$	$3 x^{2} - 9 x + 4$ $3x^2-9x+4$ $- 4$ $-4$	Subtract $4$ $4$ from both sides
$y - 4$ $y-4$	$=$ $=$	$3 x^{2} - 9 x$ $3x^2-9x$
$y - 4$ $y-4$	$=$ $=$	$3 (x^{2} - 3 x)$ $3(x^2-3x)$	Factor out $3$ $3$

Take the coefficient of the

x

$x$ term, divide it by two and then square it.

$y - 4$ $y-4$	$=$ $=$	$3 (x^{2}$ $3(x^2$ $- 3$ $-3$ $x)$ $x)$			Coefficient of the $x$ $x$ term

	$=$ $=$	$\frac{\color{#00880A}{-3}}{2}$			Divide it by $2$ $2$

		$({\frac{\color{#00880A}{-3}}{2}})^2$	$=$ $=$	$\frac{9}{4}$ $9/4$	Square

This number will make the

x

$x$ terms a perfect square.

Add and subtract $\frac{9}{4}$ $9/4$ to the grouping of

x

$x$ terms to keep the balance.

$y - 4$ $y-4$	$=$ $=$	$3 (x^{2} - 3 x)$ $3(x^2-3x)$

$y - 4$ $y-4$	$=$ $=$	$3 (x^{2} - 3 x +$ $3(x^2-3x+$ $\frac{9}{4}$ $9/4$ $-$ $-$ $\frac{9}{4}$ $9/4$ $)$ $)$	Add and subtract $\frac{9}{4}$ $9/4$

Now, transform the grouping of

x

$x$ terms into a square of a binomial.

[show cross method with two

x

$x$ ’s on the left and two

- \frac{3}{2}

$-3/2$ s on the right]

$y - 4$ $y-4$	$=$ $=$	$3 [(x - \frac{3}{2}) (x - \frac{3}{2}) - \frac{9}{4}]$ $3[(x-3/2)(x-3/2)-9/4]$

$y - 4$ $y-4$	$=$ $=$	$3 [{(x - \frac{3}{2})}^{2} - \frac{9}{4}]$ $3[(x-3/2)^2-9/4]$

Now, distribute

3

$3$ and leave

y

$y$ on the left side

$y - 4$ $y-4$	$=$ $=$	$3 [{(x - \frac{3}{2})}^{2} - \frac{9}{4}]$ $3[(x-3/2)^2-9/4]$

$y - 4$ $y-4$	$=$ $=$	$3 {(x - \frac{3}{2})}^{2} - 3 (\frac{9}{4})$ $3(x-3/2)^2-3(9/4)$	Distribute $3$ $3$

$y - 4$ $y-4$	$=$ $=$	$3 {(x - \frac{3}{2})}^{2} - \frac{27}{4}$ $3(x-3/2)^2-27/4$

$y - 4$ $y-4$ $+ 4$ $+4$	$=$ $=$	$3 {(x - \frac{3}{2})}^{2} - \frac{27}{4}$ $3(x-3/2)^2-27/4$ $+ 4$ $+4$	Add $4$ $4$ to both sides

$y$ $y$	$=$ $=$	$3 {(x - \frac{3}{2})}^{2} - \frac{11}{4}$ $3(x-3/2)^2-11/4$

Finally, the function is in vertex form

Compare the function to the general vertex form to get the vertex

$y$	$=$	$a(x-h)^2+k$

$y$	$=$	$3(x-3/2)^2-11/4$

$h$	$=$	$3/2$

	$=$	$1 1/2$

$k$	$=$	$-11/4$

	$=$	$-2 3/4$

$(1 1/2, -2 3/4)$

$- 4 x^{2} + 21 x$ $-4x^2+21x$	$=$ $=$	$x + 5$ $x+5$
$- 4 x^{2} + 21 x$ $-4x^2+21x$ $- x$ $-x$	$=$ $=$	$x + 5$ $x+5$ $- x$ $-x$	Subtract $x$ $x$ from both sides
$- 4 x^{2} + 20 x$ $-4x^2+20x$	$=$ $=$	$5$ $5$
$(- 4 x^{2} + 20 x)$ $(-4x^2+20x)$ $\div (- 4)$ $-:(-4)$	$=$ $=$	$5$ $5$ $\div (- 4)$ $-:(-4)$	Divide both sides by $- 4$ $-4$

$x^{2} - 5 x$ $x^2-5x$	$=$ $=$	$- \frac{5}{4}$ $-5/4$

$x^{2} - 5 x$ $x^2-5x$	$=$ $=$	$- \frac{5}{4}$ $-5/4$

$x^{2} - 5 x$ $x^2-5x$ $+ \frac{25}{4}$ $+25/4$	$=$ $=$	$- \frac{5}{4}$ $-5/4$ $+ \frac{25}{4}$ $+25/4$

${(x - \frac{5}{2})}^{2}$ $(x-5/2)^2$	$=$ $=$	$\frac{20}{4}$ $20/4$

${(x - \frac{5}{2})}^{2}$ $(x-5/2)^2$	$=$ $=$	$5$ $5$

$9 x^{2} + 10 x - 100$ $9x^2+10x-100$	$=$ $=$	$4 x^{2} + 15$ $4x^2+15$
$9 x^{2} + 10 x - 100$ $9x^2+10x-100$ $- 4 x^{2}$ $-4x^2$	$=$ $=$	$4 x^{2} + 15$ $4x^2+15$ $- 4 x^{2}$ $-4x^2$	Subtract $4 x^{2}$ $4x^2$ from both sides
$5 x^{2} + 10 x - 100$ $5x^2+10x-100$	$=$ $=$	$15$ $15$
$5 x^{2} + 10 x - 100$ $5x^2+10x-100$ $+ 100$ $+100$	$=$ $=$	$15$ $15$ $+ 100$ $+100$	Add $100$ $100$ to both sides
$5 x^{2} + 10 x$ $5x^2+10x$	$=$ $=$	$115$ $115$
$5 x^{2} + 10 x$ $5x^2+10x$ $\div 5$ $-:5$	$=$ $=$	$115$ $115$ $\div 5$ $-:5$	Divide both sides by $5$ $5$
$x^{2} + 2 x$ $x^2+2x$	$=$ $=$	$23$ $23$

Completing the Square 2

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2. Question

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3. Question

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4. Question

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5. Question

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Vertex Form

Quizzes

$2 x^{2} - 12 x$ $2x^2-12x$	$=$ $=$	$- 8$ $-8$
$(2 x^{2} - 12 x)$ $(2x^2-12x)$ $\div 2$ $-:2$	$=$ $=$	$- 8$ $-8$ $\div 2$ $-:2$
$x^{2} - 6 x$ $x^2-6x$	$=$ $=$	$- 4$ $-4$

$x^{2}$ $x^2$ $- 6$ $-6$ $x$ $x$	$=$ $=$	$- 4$ $-4$			Coefficient of the middle term
$- 6$ $-6$ $\div 2$ $-:2$	$=$ $=$	$- 3$ $-3$			Divide it by $2$ $2$
		${(- 3)}^{2}$ $(-3)^2$	$=$ $=$	$9$ $9$	Square

${(x - 3)}^{2}$ $(x-3)^2$	$=$ $=$	$5$ $5$
$x - 3$ $x-3$	$=$ $=$	$\sqrt{5}$ $sqrt(5)$
$x - 3$ $x-3$ $+ 3$ $+3$	$=$ $=$	$\pm \sqrt{5}$ $+-sqrt(5)$ $+ 3$ $+3$	Add $3$ $3$ to both sides
$x$ $x$	$=$ $=$	$3 \pm \sqrt{5}$ $3+-sqrt5$

$x^{2}$ $x^2$ $- 5$ $-5$ $x$ $x$	$=$ $=$	$- \frac{5}{4}$ $-5/4$			Coefficient of the middle term

$- 5$ $-5$ $\div 2$ $-:2$	$=$ $=$	$- \frac{5}{2}$ $-5/2$			Divide it by $2$ $2$
		${(- \frac{5}{2})}^{2}$ $(-5/2)^2$	$=$ $=$	$\frac{25}{4}$ $25/4$	Square

$x^{2} +$ $x^2+$ $2$ $2$ $x$ $x$	$=$ $=$	$23$ $23$			Coefficient of the middle term
$2$ $2$ $\div 2$ $-:2$	$=$ $=$	$1$ $1$			Divide it by $2$ $2$
		${(1)}^{2}$ $(1)^2$	$=$ $=$	$1$ $1$	Square

${(x + 1)}^{2}$ $(x+1)^2$	$=$ $=$	$24$ $24$
$\sqrt{{(x + 1)}^{2}}$ $sqrt((x+1)^2)$	$=$ $=$	$\sqrt{24}$ $sqrt(24)$
$x + 1$ $x+1$	$=$ $=$	$\pm 2 \sqrt{6}$ $+-2sqrt6$
$x + 1$ $x+1$ $- 1$ $-1$	$=$ $=$	$\pm 2 \sqrt{6}$ $+-2sqrt6$ $- 1$ $-1$	Subtract $1$ $1$ from both sides
$x$ $x$	$=$ $=$	$- 1 \pm 2 \sqrt{6}$ $-1+-2sqrt6$