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If p,q are the roots of the quadratic equation x 2 -10rx-11s=0 and r,s are roots of x 2 -10px-11q=0,then find the value of p+q+r+s.

If p,q are the roots of the quadratic equation x2-10rx-11s=0 and r,s are roots of x2-10px-11q=0,then find the value of p+q+r+s.

Grade:11

3 Answers

PRAMAY
22 Points
9 years ago
1210
Satyajit Samal
19 Points
9 years ago
p, q are the roots of the quadratic equation x2 - 10rx – 11s = 0 .
r, s are the roots of the quadratic equation x2 – 10px – 11q = 0.
The two equations are perfectly symmetric with roots and coefficient interchanged. 
From observation of symmetry, p = r and q = s
Now, p+q = 10 r ...(1) , pq = -11 s ...(2)
Also r+s = 10p ...(3), rs = -11q ...(4)
Multiplying (2) and (4) we get pr = 121 . Since p=r , we get p = r = – 11 (since only negative values of p and r will satisfy (2) and (4) if q = s)
Now adding (1) and (3) p+q+r+s = 10 (p+r) = 10 * (– 22) = – 220 .
 
SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
Please find the answer to your question below
Givenx2-10rx-11s=0........(1) x2-10px-11q=0....(2)
As p is a root of 1 we get p2-10pr-11s=0
and r is a root of 2 so r2-10pr-11q=0
Adding above 2 eqns we get p2+r2-20pr-11(q+s)=0
(p+r)2-22*121-99(p+r)=0
So p+r=121 or -22
As for p=-22 we get p=r so reject this
Now, p+q = 10 r ...(3) , pq = -11 s ...(4)
Also r+s = 10p ...(5), rs = -11q ...(6)
Multiplying (4) and (6) we get pr = 121 .
Now adding (1) and (3) p+q+r+s = 10 (p+r) = 10 * (121) = 1210

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