 Kye

2021-09-23

Solve the following quadratic expressions by factoring. First write the expressions in completely factored form. Then write the real numer solutions.
[Hint: Remember to use proper notation when writing the real number solutions.For example: if the solutions are $x=1$ and $x=4$, write $x=1,4$.
1) ${x}^{2}+9x+20=0$
Factored form - ?
Real number solutions - ?
2) ${x}^{2}-x-72=0$
Factored form - ?
Real number solutions - ?
3) ${x}^{2}-36=0$
Factored form - ?
Real number solutions - ?
4) ${x}^{2}-10x+25=0$
Factored form - ?
Real number solutions - ? Yusuf Keller

1) Given: ${x}^{2}+9x+20=0$
$⇒{x}^{2}+4x+5x+20=0$
$⇒x\left(x+4\right)+5\left(x+4\right)=0$
$⇒\left(x+4\right)\left(x+5\right)=0$
Factored form
$\left(x+4\right)\left(x+5\right)$Real number solution
$⇒\left(x+4\right)\left(x+5\right)=0$
$⇒x=-4,x=-5$
$x=-4,-5$
2) $⇒{x}^{2}-x-72=0$
$⇒{x}^{2}-9x+8x-72=0$
$⇒x\left(x-9\right)+8\left(x-9\right)=0$
$⇒\left(x-9\right)\left(x+8\right)$
Factored form
$\left(x-9\right)\left(x+8\right)$
Real number solution
$⇒\left(x-9\right)\left(x+8\right)=0$
$⇒x=9,x=-8$
$x=9,-8$
3) ${x}^{2}-36=0$
$⇒{x}^{2}=36$
$⇒x=\sqrt{36}$
$⇒x=6$
Factored form.
$x=6$
Real number solution.
$x=0.6$
4) ${x}^{2}-10x+25=0$
$⇒{x}^{2}-5x-5x+25=0$
$⇒{x}^{2}-5x-5x+25=0$
$⇒x\left(x-5\right)-5\left(x-5\right)=0$
$⇒\left(x-5\right)\left(x-5\right)=0$
Factored form
$\left(x-5\right)\left(x-5\right)$
Real number solution
$\left(x-5\right)\left(x-5\right)=0$
$x=5,5$

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