Question

The quadratic equation x2 – 11(p + q)x + (10p2 + 24pq + 10q2) = 0, where p ≠ ±q has

Answer 1

Step-by-step explanation:

ANSWER

p(q−r)x

2

+q(r−p)x+r(p−q)=0

D=0∴ the root are equal

D=b

2

−4ac

⇒(q(r−p))

2

−4(p(q−r))(r(p−q)))=0

⇒q 2 (r 2 +p 2−2pr)−4((pq−pr)(pr−qr))=0

⇒q 2(r 2 +p 2−2pr)−4(p 2 qr−pq 2 r−p 2 2+pqr 2)=0

⇒q 2 r 2+p 2 q 2−2pq 2 r−4p 2 qr+4pq 2r+4p 2 r 2+4pqr 2 =0

⇒q 2r 2+p 2q 2+4p 2 r 2−4p 2 qr+2pq 2

r+4pqr 2=0

⇒(pq+qr−2pr) 2=0[∵(a+b+c) 2=a 2+b 2+c 2−2ab+2bc+2ac]

⇒pq+qr=2pr

Dividing by p=qrr1+ p1= q2

Hence , m=2

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