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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

  1. 16

  2. 8

  3. 12

  4. 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| ≥ 256\sqrt{256}

⇒ |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 ≤ 644\dfrac{64}{4}

⇒ k ≤ 16 ………(4)

From (3) and (4), we get :

⇒ k = 16.

Hence, option 1 is the correct option.

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