Question

x^ +k (2x + k-1) + 2 = 0 find value of k for which quadratic equation

Answer 1

Your question is not complete

But as far as I know the question is this....
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Answer 2

Question:

Find the value of k for which the quadratic equation x² + k(2x + k - 1) + 2 = 0 has equal roots.

Answer:

k = 2

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• A quadratic equation has atmost two roots .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

The given quadratic equation is ;

x² + k(2x + k - 1) + 2 = 0 .

=> x² + 2kx + k² - k + 2 = 0

Clearly , we have ;

a = 1

b = 2k

c = k² - k + 2

We know that ,

The quadratic equation will have real and equal roots if its discriminant is zero .

=> D = 0

=> (2k)² - 4•1•(k²-k+2) = 0

=> 4k² - 4•(k²-k+2) = 0

=> 4•[k² - (k²-k+2)] = 0

=> k² - k² + k - 2 = 0

=> k - 2 = 0

=> k = 2

Hence,

The required values of k is 2 .