Question

find the value of k for which x ²- 8 x + K =0, will have real roots?
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Answer 1

Step-by-step explanation:

ax²+bx+c = 0 will have real root only when b² - 4ac >= 0

So x² + kx + 64= 0 will have real roots only when k² - 4(1)(64) > 0

k² - 256 >= 0

k² >= 256

k >= 16 or k <= - 16 (i)

Similarly x² - 8x + k = 0 will have real solution only when

64 - 4k >= 0

64 >= 4k

16 >= k

k <= 16 (ii)

From (i) and (ii)

k = 16 is the only possible solution when both these equation have real root

x²+16x+64 = 0

(x + 8)² = 0

x = - 8

x²-8x+16 =0

(x - 4)² = 0

x = 4

Answer 2

Step-by-step explanation:

The quadratic will have real root only when its discriminant will be greater than or equal to 0

Discriminant of given equation =b

2

−4ac =(−8)

2

−4×1×k>0

⇒64−4k>0

⇒4k<64

⇒k<16

So, the value of k should be less than 16.