Question

Find the value of K for which the quadratic equation k^2x^2-2(k-1)x+4=0 has two real equal roots

Answer 1

Answer:

The value of k is $-1,\frac{1}{3}$

Step-by-step explanation:

In the question we have been given a quadratic equation of x and it is said that for the right value of k we will get two equal roots from the quadratic equation.

We usually find the roots of a quadratic equation by factorization method or formula method but when we don't know the coefficients of the different powers of x we need to use a different formula for finding the value of k.

When two roots are real and equal, we can say that

$b^{2} -4ac=0$

We will be using the above formula to find the value of k.

$a=k^{2} \\b=-2(k-1)\\c=4$

$b^{2} -4ac=0\\(-2(k-1))^{2} -4.k^{2} .4=0\\4(k-1)^{2} -16k^{2} =0\\4(k^{2} -2k+1)-16k^{2} =0\\-12k^{2}-8k+4=0\\3k^{2} +2k-1=0\\3k^{2} +3k-k-1=0\\3k(k+1)-1(k+1)=0\\(k+1)(3k-1)=0$

k=-1,$\frac{1}{3}$

For these two values of k we get two equal roots for the equation.