Question

If the difference between the roots of $x^2+2px+q=0$ is two times the difference between the roots of $x^2+qx+\frac{p}{4}=0$, where $p\ne q$, then

• A. $p-q+1=0$
• B. $p-q-1=0$
• C. $p+q-1=0$
• D. $p+q+1=0$
• E. $q-4p+1=0$

Answer 1

The correct option is D $p+q+1=0$

Let the roots of the equation $x^2+2px+q=0$ be ${\alpha }_1$ and ${\beta }_1$ while roots of the equation $x^2+qx+\frac{p}{4}=0$ are ${\alpha }_2$ and ${\beta }_2$.

Then, ${\alpha }_1+{\beta }_1=-2p$

and ${\alpha }_1\cdot {\beta }_1=q$

and ${\alpha }_2+{\beta }_2=-q$

and ${\alpha }_2\cdot {\beta }_2=\frac{p}{4}$

Now, ${({\alpha }_1-{\beta }_1)}^2={({\alpha }_1+{\beta }_1)}^2-4{\alpha }_1{\beta }_1$

$=4p^2-4q=4(p^2-q)$

$\Rightarrow {\alpha }_1-{\beta }_1=2\sqrt{p^2-q}$

Also, ${({\alpha }_2-{\beta }_2)}^2={({\alpha }_2+{\beta }_2)}^2-4{\alpha }_2{\beta }_2$

$=q^2-4\times \frac{p}{4}=q^2-p$

$\Rightarrow {\alpha }_2-{\beta }_2=\sqrt{q^2-p}$

It is given that, $({\alpha }_1-{\beta }_1)=2({\alpha }_2-{\beta }_2)$

$\Rightarrow 2\sqrt{p^2-q}=2\sqrt{q^2-p}$

$\Rightarrow \sqrt{p^2-q}=\sqrt{q^2-p}$

$\Rightarrow p^2-q=q^2-p$

$\Rightarrow p^2-q^2+p-q=0$

$\Rightarrow (p-q)(p+q)+(p-q)=0$

$\Rightarrow (p-q)(p+q+1)=0$

$\Rightarrow p-q=0$ or $p+q+1=0$

but $p-q\ne 0$ as $p\ne q$

Therefore, $p+q+1=0$