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Question

In each of the following find the value of k for which the given value is a soluton of the given equation
(i)7x2+kx3=0,x=23(ii)x2x(a+b)+k=0,x=a(iii)kx2+2x4=0,x=2(iv)x23ax+k=0,x=a

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Solution

(i)k=16

putting x=23 in the equation we get

7×49+k×233=0

289+k×23=3

k×23=19

k=16

(ii)k = ab

putting x=a in the equation

a2a(a+b)+k=0

k=a2a2+ab=ab

(iii) k = 1

putting x=2 in the equation we get

k(2)2+2×24=0

2k=42×2

(iv) k = 2a^2\)


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