Question

For what values of k does the equation x(square) + (k - 2) x + k + 1 = 0 has equal roots.

Answer 1

If the equation x²+(k-2)X+k+1=0 has equal roots than the discriminant of the equation will be equal to zero.

Also we know that discriminant (D)=

b²-4ac

Which will be equal to zero because the given equation has equal roots.

Therefore b²-4ac=0

Or, (k-2)²-4(1)(k+1)= 0 {since here a= constant of x² which is 1, b=constant of x which is (k-2) and c=last constant term}

Or, {(k)²-2(k)(2)+(2)²}-{4(k+1)}=0

{Since (a-b)²= a²-2ab+b²}

Or, (k²-4k+4)-(4k+4)=0

Or, k²-4k+4-4k-4=0 {by opening brackets}

Or, k²=0

Or, k=√0

Therefore k=0

Therefore the required value of k is 0

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Answer 2

Answer:

$\red {Value \: of \: k } \green{= 0\:Or \:8 }$

Step-by-step explanation:

Given eqution x² + (k-2)x + k+1 = 0

Compare this with ax² + bx + c = 0 , we get

a = 1 , b = k-2 , c = k+1 ,

Discreminant (D) = 0 [ Roots are equal ]

=> b² - 4ac = 0

=> (k-2)² - 4×1×(k+1) = 0

=> k² - 4k + 4 - 4k - 4 = 0

=> k² - 8k = 0

=> k( k - 8) = 0

=> k = 0 Or k - 8 = 0

=> k = 0 Or k = 8

Therefore.,

Value of k = 0 Or k = 8

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