If the roots of the equation p(q-r)x2 + q(r-p)x + r(p-q) = 0 are equal, then show that 1/p + 1/r = 2/q
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Let a=p(q-r)=pq-rp
b=q(r-p)=qr-pq
c=r(p-q)=rp-qr
a+b+c=0
Now
D=b^2–4ac=(a+c)^2–4ac=0
(a-c)^2=0
a=c
a+b+c=0
Or 2a+b=0
2pq-2rp+qr-pq =0
pq-2rp+qr =0
2rp=q(p+r)
2/q=(p+r)/pr
b=q(r-p)=qr-pq
c=r(p-q)=rp-qr
a+b+c=0
Now
D=b^2–4ac=(a+c)^2–4ac=0
(a-c)^2=0
a=c
a+b+c=0
Or 2a+b=0
2pq-2rp+qr-pq =0
pq-2rp+qr =0
2rp=q(p+r)
2/q=(p+r)/pr
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