agreseza

2021-12-14

Solve the following quadratic equation by factorization method :

${x}^{2}+3x-({a}^{2}+a-2)=0$

sukljama2

Beginner2021-12-15Added 32 answers

Step 1

To solve the quadratic equation using factorization method.

Step 2

${x}^{2}+3x-({a}^{2}+a-2)=0$

${x}^{2}+3x-({a}^{2}+2a-a-2)=0$

${x}^{2}+3x-(a(a+2)-1(a+2))=0$

${x}^{2}+3x-\left((a+2)(a-1)\right)=0$

${x}^{2}+((a+2)-(a-1))x-\left((a+2)(a-1)\right)=0$

${x}^{2}+(a+2)x-(a-1)x-(a+2)(a-1)=0$

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

To solve the quadratic equation using factorization method.

Step 2

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

Beginner2021-12-16Added 33 answers

Step 1 Given

${a}^{2}{b}^{2}{x}^{2}+{b}^{2}x-{a}^{2}x-1=0$

Step 2 solving

$({a}^{2}{b}^{2}{x}^{2}+{b}^{2}x)-({a}^{2}x+1)=0$

$({a}^{2}x+1){b}^{2}x-({a}^{2}x+1)=0$

$({a}^{2}x+1)({b}^{2}x-1)=0$

${a}^{2}x+1=0,{b}^{2}x-1=0$

${a}^{2}x=-1,{b}^{2}x=1$

$x=\frac{1}{{a}^{2}},x=\frac{1}{{b}^{2}}$

Step 2 solving

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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