Question

Find the value of k for which the quadratic equation has equal roots kx^2-kx+1

Answer 1

If eqñ kx²-kx+1 has equal roots,
Then discriminatz D=0
b²-4ac=0
(k)²-4(k)(1)=0
k²=4k
k²/k=4
k=4

Answer 2

Question:

Find the value of k for which the quadratic equation kx² - kx + 1 = 0 has equal roots.

Answer:

k = 4

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 ;

kx² -kx + 1 = 0

Clearly , we have ;

a = k

b = -k

c = 1

We know that ,

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

=> D = 0

=> (-k)² - 4•k•1 = 0

=> k² - 4k = 0

=> k(k-4) = 0

=> k = 0 , 4

=> k = 4 (appreciate value)

Hence,

The required values of k is 4 .