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Question

If the roots of the quadratic equation p(qr)x2+q(rp)x+r(pq)=0 are equal,
Show then 1p+1r=2q=mq Find m.

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Solution

p(qr)x2+q(rp)x+r(pq)=0
D=0 the root are equal
D=b24ac
(q(rp))24(p(qr))(r(pq)))=0
q2(r2+p22pr)4((pqpr)(prqr))=0
q2(r2+p22pr)4(p2qrpq2rp2r2+pqr2)=0
q2r2+p2q22pq2r4p2qr+4pq2r+4p2r2+4pqr2=0
q2r2+p2q2+4p2r24p2qr+2pq2r+4pqr2=0
(pq+qr2pr)2=0[(a+b+c)2=a2+b2+c22ab+2bc+2ac]
pq+qr=2pr
Dividing by p=qr
1r+1p=2q Hence , m=2

1170542_1278472_ans_d65df0a00e45418081858a87be5ce9da.jpg

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