agreseza

2021-12-14

Solve the following quadratic equation by factorization method :
${x}^{2}+3x-\left({a}^{2}+a-2\right)=0$

sukljama2

Step 1
To solve the quadratic equation using factorization method.
Step 2
${x}^{2}+3x-\left({a}^{2}+a-2\right)=0$
${x}^{2}+3x-\left({a}^{2}+2a-a-2\right)=0$
${x}^{2}+3x-\left(a\left(a+2\right)-1\left(a+2\right)\right)=0$
${x}^{2}+3x-\left(\left(a+2\right)\left(a-1\right)\right)=0$
${x}^{2}+\left(\left(a+2\right)-\left(a-1\right)\right)x-\left(\left(a+2\right)\left(a-1\right)\right)=0$
${x}^{2}+\left(a+2\right)x-\left(a-1\right)x-\left(a+2\right)\left(a-1\right)=0$
x(x+(a+2))-(a-1)(x+(a+2))=0
(x+(a+2))(x-(a-1))=0
x+(a+2)=0 or x-(a-1)=0
x=-a-2, x=a-1

John Koga

Step 1 Given
${a}^{2}{b}^{2}{x}^{2}+{b}^{2}x-{a}^{2}x-1=0$
Step 2 solving
$\left({a}^{2}{b}^{2}{x}^{2}+{b}^{2}x\right)-\left({a}^{2}x+1\right)=0$
$\left({a}^{2}x+1\right){b}^{2}x-\left({a}^{2}x+1\right)=0$
$\left({a}^{2}x+1\right)\left({b}^{2}x-1\right)=0$
${a}^{2}x+1=0,{b}^{2}x-1=0$
${a}^{2}x=-1,{b}^{2}x=1$
$x=\frac{1}{{a}^{2}},x=\frac{1}{{b}^{2}}$

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