Question

X^2-4x+(k+1) equal root and real

Answer 1

x²-4x + k+1

Here,

a = 1

b = -4

c = k+1

Since the equation has equal roots,

D = b²-4ac = 0

b² = 4ac

(-4)² = 4*1*(k+1)

16 = 4(k+1)

k+1 = 4

k = 3

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Answer 2

Answer:

k=3.75

Step-by-step explanation:

Let p(x) = $x^2-4x+(k+1)$

Given that the roots are real and equal.

∴D or $b^2-4ac$ = 0

$(-4)^2-4(1)(k+1)=0$

16-4(k+1) = 0

16-4k-1 = 0

15-4k = 0

-4k = -15

k= -15/-4

k=15/4

k=3.75

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