Question

find the value of k for which the give equation has real and equal root :(k+1) x²-2(k-1)x+1=0​

Answer 1

Answer:

The value of k is 3

Given:

(k+1)x×2-2(k-1)x+1=0(k+1)x×2−2(k−1)x+1=0

To find:

The value of k

Solution:

We know that,

For a quadratic equation x^{2}+b x+c=0x

2

+bx+c=0 , if the roots are real and equal then

Discriminant b^{2}-4ac=0 \rightarrow(1)b

2

−4ac=0→(1)

Now, in an equation (k+1) x \text { times } 2-2(k-1) x+1=0(k+1)x times 2−2(k−1)x+1=0

a = k+1, b= -2(k-1), c=1a=k+1,b=−2(k−1),c=1

Given that roots are real and equal

Therefore,

[-2(k-1)]^{2}-4(k+1)(1]=0[−2(k−1)]

2

−4(k+1)(1]=0

Now, the above equation can be written as,

4\left(k^{2}-2 k+1\right)-4 k-4=04(k

2

−2k+1)−4k−4=0

On simplifying the above equation, we get

4\left(k^{2}-2 k+1\right)-4 k-4=04(k

2

−2k+1)−4k−4=0

Now, +4 and -4 get cancelled,

\begin{gathered}\begin{array}{c}{4 k^{2}-12 k=0} \\\\ {4 k^{2}=12 k} \\\\ {k=3}\end{array}\end{gathered}

4k

2

−12k=0

4k

2

=12k

k=3

Explanation:

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Answer 2

Explanation:

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