Question

If the quadratic equation ax2_4ax +2a +1=0 has Repeated(equal) roots, a=

Answer 1

Solved!!! Any query ask

Answer 2

Given,

An equation ax²-4ax+2a+1=0.

To find,

To find the value of a for which equation ax²-4ax+2a+1=0 has equal roots. Solution,

For the quadratic equation having equal roots, their Discriminant is always equal to zero.

Suppose an equation is of form mx²+nx+c=0 where m, n, and c are constant.

Discriminant (D) is calculated by D = n²-4mc.

In the given question,

m = a, n = -4a, c = 2a+1

Hence D = (-4a)²-4(a)(2a+1) = 0

⇒   D = 16a²-8a²-4a = 0

⇒   D = 8a²-4a = 0

⇒   4a(2a-1) = 0

⇒   a = 0 and a = 1/2

Hence the value of a for which equation ax²-4ax+2a+1=0  has equal roots are a = 0 and a = 1/2.