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Let p,q and r be real numbers (p≠q,r≠0, such that the roots of the equation 
are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to
are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to- a)p2+q2
- b)
- c)2(p2+q2)
- d)p2+q2+r2
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Let p,q and r be real numbers (p≠q,r≠0, such that the roots of t...
Given quadratic equation is 
Let α and β be the roots of given equation.
⇒(2x+p+q)r=(x+p)(x+q)
⇒x2+(p+q−2r)x+pq−pr−qr=0
Now, sum of roots (α+β)= -b/a =−(p+q−2r)
⇒−(p+q−2r)=0 (∵Given that roots are equal in magnitude and opposite in sign)
⇒p+q=2r ...(1)
Product of roots (αβ)= c/a =pq−pr−qr
Now, α2+β2=(α+β)2−2αβ
=0−2[pq−pr−qr]=−2pq+2r(p+q)
=−2pq+(p+q)2=p2+q2 (∵ from (1)
=p2+q2
Let α and β be the roots of given equation.
⇒(2x+p+q)r=(x+p)(x+q)
⇒x2+(p+q−2r)x+pq−pr−qr=0
Now, sum of roots (α+β)= -b/a =−(p+q−2r)
⇒−(p+q−2r)=0 (∵Given that roots are equal in magnitude and opposite in sign)
⇒p+q=2r ...(1)
Product of roots (αβ)= c/a =pq−pr−qr
Now, α2+β2=(α+β)2−2αβ
=0−2[pq−pr−qr]=−2pq+2r(p+q)
=−2pq+(p+q)2=p2+q2 (∵ from (1)
=p2+q2
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Let p,q and r be real numbers (p≠q,r≠0, such that the roots of the equationare equal in magnitude but opposite in sign, then the sum of squares ofthese roots is equal toa)p2+q2b)c)2(p2+q2)d)p2+q2+r2Correct answer is option 'A'. Can you explain this answer?
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