Find k so that the quadratic equation (k+1) x^-2(k+1=0has two equal root
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Secondary SchoolMath 5+3 pts
If the quadratic equation (k+1)x*2-2(k-1)X+1=0 have real and equal roots then find the value of k
Report by Dishapakhi 03.12.2017
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Aryanbansi1111 Ambitious
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Gratefuljarette Ace
The value of k is 3
Given:
(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=0, if the roots are real and equal then
Discriminant b^{2}-4ac=0 \rightarrow(1)
Now, in an equation (k+1) x \text { times } 2-2(k-1) x+1=0
a = 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
Now, the above equation can be written as,
4\left(k^{2}-2 k+1\right)-4 k-4=0
On simplifying the above equation, we get
4\left(k^{2}-2 k+1\right)-4 k-4=0
Now, +4 and -4 get cancelled,
\begin{array}{c}{4 k^{2}-12 k=0} \\\\ {4 k^{2}=12 k} \\\\ {k=3}\end{array}