KEAM 2015 Math Paper

Let the roots of the equation x2+2px+q=0 be α1 and β1 while roots of the equation x2+qx+‌

p

4

=0
are α2 and β2
Then, ‌‌α1+β1=−2p
and ‌‌α1⋅β1=q
and ‌‌α2+β2=−q
and ‌‌α2⋅β2=‌

p

4

Now, (α1−β1)2=(α1+β1)2−4α1β1
=4p2−4q=4(p2−q)
⇒‌‌α1−β1=2√p2−q
Also, (α2−β2)2=(α2+β2)2−4α2β2
=q2−4×‌

p

4

=q2−p
⇒‌‌α2−β2=√q2−p
It is given that, (α1−β1)=2(α2−β2)
⇒2√p2−q=2√q2−p
⇒√p2−q=√q2−p
⇒‌‌p2−q=q2−p
⇒‌‌p2−q2+p−q=0
⇒(p−q)(p+q)+(p−q)=0
⇒‌‌(p−q)(p+q+1)=0
⇒‌‌p−q=0 or p+q+1=0
but p−q≠0 as p≠q
∴p+q+1=0