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
(a)4
(b)8
(c)12
(d)16

Answer 1

SOLUTION :  

Option (d) is correct : 16

Given : x² + kx + 64 = 0 …………(1)

and x² - 8x + k = 0…………..(2)

On comparing the given equation with ax² + bx + c = 0  

Let D1 & D2 be the discriminants of the two given equations .

For eq 1 :  

Here, a = 1 , b =  k , c = 64

D(discriminant) = b² – 4ac

D1 = (k)² - 4 × 1 × 64

D1 = k² - 256……….(3)

For eq 2 :  

x² - 8x + k = 0

Here, a = 1  ,b =  -8, c = k

D(discriminant) = b² – 4ac

D2 = (-8)² - 4 × 1 × k

D2 = 64 - 4k …………….. (4)

Given : Roots are real for both the Given equations i.e D ≥ 0.

D1 ≥ 0  

k² - 256 ≥ 0  

[From eq 3]

k² ≥ 256

k ≥ 16 …………..(5)

D2 ≥ 0

64 - 4k ≥ 0

64  ≥ 4k

16  ≥ k

k  ≤ 16 …………….(6)

k = 16   [From eq 5 & 6 ]  

Hence, the value of k is 16 .

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Let D₁ and D₂ be the discriminant of the given equations and these will have equal roots only if D₁, D₂ ≥ 0.

Or, if D₁ = (k² -4 × 64) ≥ 0 and D₂ = (64 - 4k) ≥ 0.

Or, if k² ≥ 256 and 4k ≤ 64.

Or, if k ≥ 16 and k ≤ 16.

Or, k = 16.

Thus, the positive value of/for k is 16.