Completing the Square 1

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Question 1 of 5

Solve by completing the square

$x^2-12x=-22$

1. $6+-sqrt(14)$
2. $-6+-sqrt(14)$
3. $6+-sqrt(10)$
4. $-6+-sqrt(10)$

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Completing the square is done by taking the coefficient of

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

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

$x^2$ $-12$ $x$	$=$	$-22$			Coefficient of the middle term
$-12$ $-:2$	$=$	$-6$			Divide it by $2$
		$(-6)^2$	$=$	$36$	Square

This number will make the left side a perfect square.

Add $36$ to both sides of the equation to keep the balance.

$x^2-12x$	$=$	$-22$
$x^2-12x$ $+36$	$=$	$-22$ $+36$	Add $36$ to both sides
$x^2-12x+36$	$=$	$14$

Now, transform the left side into a square of a binomial by factoring or using cross method.

$(x-6)(x-6)$	$=$	$14$
$(x-6)^2$	$=$	$14$

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

$x$ .

$(x-6)^2$	$=$	$14$
$sqrt((x-6)^2)$	$=$	$sqrt14$	Take the square root
$x-6$	$=$	$+-sqrt(14)$	Square rooting a number gives a plus and minus solution
$x-6$ $+6$	$=$	$+-sqrt(14)$ $+6$	Add $6$ to both sides
$x$	$=$	$6+-sqrt(14)$	Simplify

The roots can also be written individually

$=$

$6+\sqrt14$

$=$

$6-\sqrt14$

$x=6+-sqrt(14)$

Question 2 of 5

2. Question
Solve by completing the square

$x^2+6x+13=0$
- 1. $x=-2,2$
- 2. $x=-3,7$
- 3. $x=-6,0$
- 4. No real roots
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Completing the square is done by taking the coefficient of $x$ , halving it and then squaring it. Then we add the new value to both sides of the equation.

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

$x^2+$ $6$ $x+13$ $=$ $0$ Coefficient of the middle term

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

$(3)^2$ $=$ $9$ Square

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

Add and subtract $9$ to the left side of the equation to keep the balance, then form a square of a binomial

$x^2+6x+13$ $=$ $0$

$x^2+6x$ $+9-9$ $+13$ $=$ $0$

$(x+3)^2-9+13$ $=$ $0$

$(x+3)^2+4$ $=$ $0$

Move the constant to the right

$(x+3)^2+4$ $=$ $0$

$(x+3)^2+4$ $-4$ $=$ $0$ $-4$ Subtract $4$ from both sides

$(x+3)^2$ $=$ $-4$

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

$(x+3)^2$ $=$ $-4$

$sqrt((x+3)^2)$ $=$ $sqrt(-4)$

Remember that a negative value inside a surd will give out imaginary roots. Therefore, this equation has no real roots

No real roots

Question 3 of 5

Solve by completing the square

$x^2+3x-6=0$

1. $(-3+-sqrt33)/2$
2. $(3+-sqrt30)/2$
3. $(+-sqrt33)/2$
4. $(-2+-sqrt33)/3$

Question 4 of 5

4. Question
Solve by completing the square

$k^2-4k=2k+18$
- 1. $1+-sqrt3$
- 2. $+-3sqrt2$
- 3. $-3+-2sqrt5$
- 4. $3+-3sqrt3$
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Completing the square is done by taking the coefficient of $x$ , halving it and then squaring it. Then we add the new value to both sides of the equation.

Move $k$ terms to the left

$k^2-4k$ $=$ $2k+18$

$k^2-4k$ $-2k$ $=$ $2k+18$ $-2k$ Subtract $2k$ from both sides

$k^2-6k$ $=$ $18$

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

$k^2$ $-6$ $k$ $=$ $18$ Coefficient of the middle term

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

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

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

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

$k^2-6k$ $=$ $18$

$k^2-6k$ $+9$ $=$ $18$ $+9$

$(x-3)^2$ $=$ $27$

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

$(x-3)^2$ $=$ $27$

$sqrt((x-3)^2)$ $=$ $sqrt(27)$

$x-3$ $=$ $+-3sqrt3$

$x-3$ $+3$ $=$ $+-3sqrt3$ $+3$ Add $3$ to both sides

$x$ $=$ $3+-3sqrt3$

The roots can also be written individually

$=$ $3+3sqrt3$

$=$ $3-3sqrt3$

$3+-3sqrt3$

Question 5 of 5

Solve by completing the square

$u^2+1.8u-2.2=0$

1. $-0.9+-sqrt(3.01)$
2. $0.5+-sqrt(2)$
3. $-2.1+-sqrt(3.5)$
4. $0.95+-sqrt(1.02)$

$(x+3/2)^2-33/4$	$=$	$0$

$(x+3/2)^2-33/4$ $+33/4$	$=$	$0$ $+33/4$	Add $33/4$ to both sides

$(x+3/2)^2$	$=$	$33/4$

$(u+0.9)^2-3.01$	$=$	$0$
$(u+0.9)^2-3.01$ $+3.01$	$=$	$0$ $+3.01$	Add $3.01$ to both sides
$(u+0.9)^2$	$=$	$3.01$

$(u+0.9)^2$	$=$	$3.01$
$sqrt((u+0.9)^2)$	$=$	$sqrt(3.01)$
$u+0.9$	$=$	$+-sqrt(3.01)$
$u+0.9$ $-0.9$	$=$	$+-sqrt(3.01)$ $-0.9$	Subtract $0.9$ from both sides
$u$	$=$	$-0.9+-sqrt(3.01)$

Completing the Square 1

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$x^2+$ $6$ $x+13$	$=$	$0$			Coefficient of the middle term
$6$ $-:2$	$=$	$3$			Divide it by $2$
		$(3)^2$	$=$	$9$	Square

$x^2+6x+13$	$=$	$0$
$x^2+6x$ $+9-9$ $+13$	$=$	$0$
$(x+3)^2-9+13$	$=$	$0$
$(x+3)^2+4$	$=$	$0$

$(x+3)^2+4$	$=$	$0$
$(x+3)^2+4$ $-4$	$=$	$0$ $-4$	Subtract $4$ from both sides
$(x+3)^2$	$=$	$-4$

$x^2+$ $3$ $x-6$	$=$	$0$			Coefficient of the middle term

$3$ $-:2$	$=$	$3/2$			Divide it by $2$

		$(3/2)^2$	$=$	$9/4$	Square

$x^2+3x-6$	$=$	$0$

$x^2+3x$ $+9/4-9/4$ $-6$	$=$	$0$

$(x+3/2)^2-9/4-6$	$=$	$0$

$(x+3/2)^2-33/4$	$=$	$0$

$(x+3/2)^2$	$=$	$33/4$

$sqrt((x+3/2)^2)$	$=$	$sqrt(33/4)$

$x+3/2$	$=$	$(+-sqrt33)/2$

$x+3/2$ $-3/2$	$=$	$(+-sqrt33)/2$ $-3/2$	Subtract $3/2$ from both sides
$x$	$=$	$(-3+-sqrt33)/2$

$k^2-4k$	$=$	$2k+18$
$k^2-4k$ $-2k$	$=$	$2k+18$ $-2k$	Subtract $2k$ from both sides
$k^2-6k$	$=$	$18$

$k^2$ $-6$ $k$	$=$	$18$			Coefficient of the middle term
$-6$ $-:2$	$=$	$-3$			Divide it by $2$
		$(-3)^2$	$=$	$9$	Square

$k^2-6k$	$=$	$18$
$k^2-6k$ $+9$	$=$	$18$ $+9$
$(x-3)^2$	$=$	$27$

$(x-3)^2$	$=$	$27$
$sqrt((x-3)^2)$	$=$	$sqrt(27)$
$x-3$	$=$	$+-3sqrt3$
$x-3$ $+3$	$=$	$+-3sqrt3$ $+3$	Add $3$ to both sides
$x$	$=$	$3+-3sqrt3$

$u^2+$ $1.8$ $u-2.2$	$=$	$0$			Coefficient of the middle term
$1.8$ $-:2$	$=$	$0.9$			Divide it by $2$
		$(0.9)^2$	$=$	$0.81$	Square

$u^2+1.8u-2.2$	$=$	$0$
$u^2+1.8u$ $+0.81-0.81$ $-2.2$	$=$	$0$
$(u+0.9)^2-0.81-2.2$	$=$	$0$
$(u+0.9)^2-3.01$	$=$	$0$