Question

Solve the following quadratic equation by factorization, the roots are $4,-7$
$\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1};x\ne 1,-1,\frac{1}{4}$

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1}$

$\Rightarrow$ $\frac{3(x-1)+4(x+1)}{(x+1)(x-1)}=\frac{29}{4x-1}$

$\Rightarrow$ $\frac{3x-3+4x+4}{x^2-1}=\frac{29}{4x-1}$

$\Rightarrow$ $\frac{7x+1}{x^2-1}=\frac{29}{4x-1}$

$\Rightarrow$ $(7x+1)(4x-1)=29(x^2-1)$

$\Rightarrow$ $28x^2+4x-7x-1=29x^2-29$

$\Rightarrow$ $28x^2-3x-1=29x^2-29$

$\Rightarrow$ $x^2+3x-28=0$

$\Rightarrow$ $x^2+7x-4x-28=0$

$\Rightarrow$ $x(x+7)-4(x+7)=0$

$\Rightarrow$ $(x+7)(x-4)=0$

$\Rightarrow$ $x+7=0$ and $x-4=0$

$\therefore$ $x=-7$ and $x=4$

$\therefore$ We can see, roots given in question are correct.