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# If c, d are zeroes of x2-10x-11b and a , b are zeroes of x2-10cx-11d then value of a+b+c+d= ? A) 1210 B)-1 C)2530 D)-11

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Solution

## We know for sure, from the properties of the roots that: a+b=10ca+b=10c c+d=10ac+d=10a a.b=−11da.b=−11d c.d=−11bc.d=−11b Which implies a.c=121a.c=121 and (a+c)=(b+d)/9(a+c)=(b+d)/9 As aa is a root of the first equation, a2−10ac−11d=0a2−10ac−11d=0 and as cc is a root of the second equation, c2−10ac−11b=0c2−10ac−11b=0 Adding both equations, we get: ⟹a2+c2−2∗10ac−11(b+d)=0⟹a2+c2−2∗10ac−11(b+d)=0 ⟹a2+c2−2∗10ac−2∗ac+2∗ac−11∗9(a+c)=0⟹a2+c2−2∗10ac−2∗ac+2∗ac−11∗9(a+c)=0 ⟹(a+c)2−22(ac)−99(a+c)=0⟹(a+c)2−22(ac)−99(a+c)=0 Substituting the value of ac=121ac=121, we get a quadratic equation where (a+c)(a+c)is the variable. ⟹(a+c)2−99(a+c)−2662=0⟹(a+c)2−99(a+c)−2662=0 Let X=(a+c)X=(a+c). ⟹X2−99X−2662=0⟹X2−99X−2662=0 ⟹X=(a+c)=−22or121⟹(a+b+c+d)=−220or1210

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