Question

solving quadratic equations by factoring
solve each equation by factoring.

1. $(k+1)(k-5)=0$
2. $(a+1)(a+2)=0$
3. $(4k+5)(k+1)=0$
4. $(2m+3)(4m+3)=0$
5. $x^{2}-11x+19=-5$
6. $n^{2}+7n+15=5$
7. $n^{2}-10n+22=-2$
8. $n^{2}+3n-12=6$
9. $6n^{2}-18n-18=6$
10. $7r^{2}-14r=-7$

(handwritten solution attempts are present but omitted as they are not part of the original problem statements)

Explanation:

Step1: Set each factor to 0

For $(k+1)(k-5)=0$, we get:
$k+1=0$ or $k-5=0$

Step2: Solve for $k$

$k=-1$ or $k=5$

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Step1: Set each factor to 0

For $(a+1)(a+2)=0$, we get:
$a+1=0$ or $a+2=0$

Step2: Solve for $a$

$a=-1$ or $a=-2$

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Step1: Set each factor to 0

For $(4k+5)(k+1)=0$, we get:
$4k+5=0$ or $k+1=0$

Step2: Solve for $k$

$4k=-5 \implies k=-\frac{5}{4}$, or $k=-1$

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Step1: Set each factor to 0

For $(2m+3)(4m+3)=0$, we get:
$2m+3=0$ or $4m+3=0$

Step2: Solve for $m$

$2m=-3 \implies m=-\frac{3}{2}$, or $4m=-3 \implies m=-\frac{3}{4}$

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Step1: Rewrite to standard form

$x^2-11x+19=-5 \implies x^2-11x+24=0$

Step2: Factor the quadratic

$x^2-11x+24=(x-3)(x-8)=0$

Step3: Set factors to 0 and solve

$x-3=0 \implies x=3$, or $x-8=0 \implies x=8$

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Step1: Rewrite to standard form

$n^2+7n+15=5 \implies n^2+7n+10=0$

Step2: Factor the quadratic

$n^2+7n+10=(n+2)(n+5)=0$

Step3: Set factors to 0 and solve

$n+2=0 \implies n=-2$, or $n+5=0 \implies n=-5$

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Step1: Rewrite to standard form

$n^2-10n+22=-2 \implies n^2-10n+24=0$

Step2: Factor the quadratic

$n^2-10n+24=(n-4)(n-6)=0$

Step3: Set factors to 0 and solve

$n-4=0 \implies n=4$, or $n-6=0 \implies n=6$

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Step1: Rewrite to standard form

$n^2+3n-12=6 \implies n^2+3n-18=0$

Step2: Factor the quadratic

$n^2+3n-18=(n+6)(n-3)=0$

Step3: Set factors to 0 and solve

$n+6=0 \implies n=-6$, or $n-3=0 \implies n=3$

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Step1: Rewrite to standard form

$6n^2-18n-18=6 \implies 6n^2-18n-24=0$

Step2: Simplify the equation

Divide by 6: $n^2-3n-4=0$

Step3: Factor the quadratic

$n^2-3n-4=(n-4)(n+1)=0$

Step4: Set factors to 0 and solve

$n-4=0 \implies n=4$, or $n+1=0 \implies n=-1$

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Step1: Rewrite to standard form

$7r^2-14r=-7 \implies 7r^2-14r+7=0$

Step2: Simplify the equation

Divide by 7: $r^2-2r+1=0$

Step3: Factor the quadratic

$r^2-2r+1=(r-1)^2=0$

Step4: Solve for $r$

$r-1=0 \implies r=1$ (repeated root)

Answer:

1. $k=-1$ or $k=5$
2. $a=-1$ or $a=-2$
3. $k=-\frac{5}{4}$ or $k=-1$
4. $m=-\frac{3}{2}$ or $m=-\frac{3}{4}$
5. $x=3$ or $x=8$
6. $n=-2$ or $n=-5$
7. $n=4$ or $n=6$
8. $n=-6$ or $n=3$
9. $n=4$ or $n=-1$
10. $r=1$