Question

find the value of k for which the quadratic equation 2x^2kx + k = 0 had equal roots

Answer 1

Solution:
For equal roots : b^2-4ac =0
hence, k^2 - 4×2×k =0
● k^2 - 8k =0
● k(k-8) =0
● k =8
Thanks for asking

Answer 2

Question:

Find the value of k for which the quadratic equation 2x² -kx + k= 0 has equal roots.

Answer:

k = 0 or 8

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 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 ;

2x² - kx + k = 0 .

Clearly , we have ;

a = 2

b = -k

c = k

We know that ,

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

=> D = 0

=> (-k)² - 4•2•k = 0

=> k² - 8k = 0

=> k•(k-8) = 0

=> k = 0 or 8

Hence,

The required values of k are 0 and 8 .