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KEAM 2015 Math Paper

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Question : 33 of 120

Marks: +1, -0

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,

p\neq q,

then

$p - q + 1=0$
$p - q - 1=0$
$p + q - 1=0$
$p + q + 1=0$
$p - 4q + 1=0$

Solution:

Let the roots of the equation

x^{2}+2 p x+q=0

be

\alpha_{1}

and

\beta_{1}

while roots of the equation

x^{2}+q x+\;\frac{p}{4}=0

are

\alpha_{2}

and

\beta_{2}

Then,

\;\; \alpha_{1}+\beta_{1}=-2 p

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}

=4 p^{2}-4 q=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 \neq 0

as

p \neq q

\therefore p+q+1=0

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