Question

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

Answer 1

I hope this may help you

Answer 2

Solution:

As the question says, the roots of the equation are equal. Let us take the roots as$\alpha$, $\beta$

Now we can say that $\alpha=\beta=m$

Therefore, the equation is $x^{2}+k 2 x+k^{2}-k+2$which means if put according to a $x^{2}+b x+c$,

Then the sum of root is, $\alpha+\beta=-\frac{b}{a}b$

that is$2 m=-\frac{2 k}{1}$ and

$\alpha \beta=\frac{c}{a}=\frac{k^{2}-k+2}{1}.$

Therefore,  

$\boldsymbol{m}=\boldsymbol{k} \& \boldsymbol{m}^{2}=\boldsymbol{k}^{2}-\boldsymbol{k}+\mathbf{2}$

Put the value of m = k in$m^{2}=k^{2}-k+2$ we get

$k^{2}=k^{2}-k+2$

k=2  

Therefore, the value of k = 2.