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Question
The positive value of k for which the equation x2 + kx + 64 = 0 and x2 − 8x + k = 0 will both have real roots, is
Options
4
8
12
16
Solution
The given quadric equation are x2 + kx + 64 = 0 and x2 − 8x + k = 0 roots are real.
Then find the value of a.
Here, x2 + kx + 64 = 0 ….. (1)
x2 − 8x + k = 0 ….. (2)
`a_1 = 1,b_1 = k and ,c_1 = 64`
`a_2 = 1,b_2 = -8 and ,c_2 = k`
As we know that `D_1 = b^2 - 4ac`
Putting the value of `a_1 = 1,b_1 = k and ,c_1 = 64`
`=(k)^2 - 4 xx 1 xx 64`
`= k^2 - 256`
The given equation will have real and distinct roots, if D >0
`k^2 - 256 = 0`
`k^2 = 256`
`k = sqrt256`
` k = ± 16`
Therefore, putting the value of k = 16 in equation (2) we get
` x^2 - 8x + 16 = 0`
`(x - 4)^2 = 0`
x - 4 = 0
x = 4
The value of k = 16 satisfying to both equations
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