Question

using factorisation method solve this quadratic equation

NO SPAM

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

Answer 1

Step-by-step explanation:

Answer:

\green {x = \frac{a+b}{a}\:Or\:x = \frac{a+b}{b} }

Step-by-step explanation:

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

\implies abx^{2}- (a+b)^{2} (x-1) = 0

\implies abx^{2} -(a+b)^{2}x + (a+b)^{2}=0

/* Divide each term by ab ,we get

\implies x^{2} -\frac{(a+b)}{ab} x + \frac{(a+b)^{2}}{ab}=0

\implies x^{2} - \frac{(a+b)}{a} x - \frac{(a+b)}{b} x + \frac{(a+b)^{2}}{ab} = 0

/* Splitting the middle term, we get

\implies x\left( x - \frac{a+b}{a}\right) -\frac{(a+b)}{b} \left( x - \frac{a+b}{a}\right)=0

\implies \left( x - \frac{a+b}{a}\right)\left( x - \frac{a+b}{b}\right) = 0

\implies x - \frac{a+b}{a}=0 \:Or \: x - \frac{a+b}{b} = 0

\therefore x = \frac{a+b}{a}\:Or\:x = \frac{a+b}{b}