11a+b=10c
c+d=10a
=> 11(a-c)=d-b.........(1)
Put a and c in their respective eqtns:
a^2-10ac-11d=0
c^2-10ac-11b=0
=> (a-c)(a+c) =11(d-b).....(2)
From (1) and (2) :
(a-c)(a+c)=121(a-c)
=>a+c=121
Now a+b+c+d=10(a+c)=10*121=1210...(Ans)
if roots of the equation x2-10cx-11d=0 are a,b and those of x2-10ax-11b=0 are c,d,then find the value of (a+b+c+d)?
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2 Answers
11
71As a + b = 10c and c + d = 10a
ab = −11d , cd = −11b
⇒ ac = 121 and (b + d) = 9(a + c)
a2 − 10ac − 11d = 0
c2 − 10ac − 11b = 0
⇒ a2 + c2 − 20ac − 11(b + d) = 0
⇒ (a + c)2 − 22(121) − 11 × 9(a + c) = 0
⇒ (a + c) = 121 or −22 (rejected)
∴ a + b + c + d = 1210.
