Mathematics
The positive value of k for which the equations x2 + kx + 64 = 0 and x2 - 8x + k = 0 will both have real roots, is
16
8
12
4
Quadratic Equations
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Answer
Given,
x2 + kx + 64 = 0 ….(1)
x2 - 8x + k = 0 ….(2)
Comparing x2 + kx + 64 = 0 with ax2 + bx + c = 0 we get,
a = 1, b = k and c = 64.
We know that,
Discriminant (D) = b2 - 4ac
= (k)2 - 4 × (1) × (64)
= k2 - 256
Since equations has real roots,
⇒ D ≥ 0
⇒ k2 - 256 ≥ 0
⇒ k2 ≥ 256
⇒ |k| ≥
⇒ |k| ≥ 16
⇒ 16 ≤ k ≤ -16
Taking only positive value,
⇒ k ≥ 16 ………(3)
Comparing x2 - 8x + k = 0 with ax2 + bx + c = 0 we get,
a = 1, b = -8 and c = k.
We know that,
Discriminant (D) = b2 - 4ac
= (-8)2 - 4 × (1) × (k)
= (64) - 4k
Since equations has real roots,
⇒ D ≥ 0
⇒ 64 - 4k ≥ 0
⇒ 4k ≤ 64
⇒ k ≤
⇒ k ≤ 16 ………(4)
From (3) and (4), we get :
⇒ k = 16.
Hence, option 1 is the correct option.
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