Question

Find values of k for which the quadratic equation x2+k(2x+k-1)+2 which has equal roots

Answer 1

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

for equal roots d= 0
d= b^2 -4ac

= 4k^2 - 4k^2 + 4k - 8 = 0
= 4k - 8 = 0
= 4k = 8
= k = 8/4
= k= 2

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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 .

• 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 equal to zero .

=> D = 0

=> (2k)² - 4•1•(k²-k+2) = 0

=> 4k² - 4(k²-k+2) = 0

=> 4(k²-k²+k-2) = 0

=> k - 2 = 0

=> k = 2

Hence,

The required values of k is 2 .