Algebra> If p,q are the roots of the quadratic equ...

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.

Shubhanshu Singh , 10 Years ago

Grade 11

3 Answers

PRAMAY

Last Activity: 10 Years ago

1210

Satyajit Samal

Last Activity: 10 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

Last Activity: 10 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