Question

Let $x_1$ and $x_2$ be the roots of the quadratic equation $x^2+px+q=0$.
If $x_1=\frac{x_2+4}{2x_2-1}$, then the value of $(2q+p)$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$1$

Sum of roots $=x_1+x_2=-p$

product of roots $=x_1{x}_2=q$

$\therefore x_1+x_2=\frac{x_2+4}{2x_2-1}+x_2=-p$

$\therefore \frac{x_2+4+x_2}{2x_2-1}(2x_2-1)=-p$

$\therefore \frac{2x_2^2+4}{2x_2-1}=-p$

Now $\frac{x_2^2+4x_2}{2x_2-1}=q$

$\therefore 2q=\frac{2x_2^2+8x_2}{2x_2-1}$

$\therefore 2q+p=\frac{2x_2^2+8x_2+2x_2^2}{2x_2-1}+4$

$\therefore 2q+p=\frac{4x_2^2+8x_2+4}{2x_2-1}$

$\therefore 2q+p=1$