Question

Determine the positive value of K for whoch the equation x²-kx+64=0and k²-8x+k=0will vorh have real root.

Answer 1

Answer:

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Step-by-step explanation:

For x²-8x+k=0, D=(-8)²-4*1*k=64–4k. For real roots, 64–4k>=0, which is true for k<=16. So, the common point for k>=16 and k <=16 is only k=16.

Answer 2

Answer

x^2-kx+64=0

b^2-4ac=0

a=1, b=k, c=64

k^2 - 4×1×64 =0

k^2 - 256 =0

k^2 = 256

k = +-√256

k=+-16

k=16

k^2-8x+k=0

b^2-4ac=0

a=1, b=-8, c=k

8^2 -4×1×k=0

64-4k=0

-4k = -64

k= 64/4

k=16