If x2+k(4x+k−1)+2=0 has equal roots, then k =
Question:

If $x^{2}+k(4 x+k-1)+2=0$ has equal roots, then $\mathrm{k}=$

(a) $-\frac{2}{3}, 1$

(b) $\frac{2}{3},-1$

(c) $\frac{3}{2}, \frac{1}{3}$

(d) $-\frac{3}{2},-\frac{1}{3}$

Solution:

The given quadric equation is $x^{2}+k(4 x+k-1)+2=0$, and roots are equal

Then find the value of k.

$x^{2}+k(4 x+k-1)+2=0$

$x^{2}+4 k x+\left(k^{2}-k+2\right)=0$

Here, $a=1, b=4 k$ and, $c=k^{2}-k+2$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=1, b=4 k$ and, $c=k^{2}-k+2$

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

$=16 k^{2}-4 k^{2}+4 k-8$

$=12 k^{2}+4 k-8$

$=4\left(3 k^{2}+k-2\right)$

The given equation will have real and distinct roots, if $D=0$

$4\left(3 k^{2}+k-2\right)=0$

$3 k^{2}+k-2=0$

$3 k^{2}+3 k-2 k-2=0$

$3 k(k+1)-2(k+1)=0$

$(k+1)(3 k-2)=0$

Therefore, the value of $k=\frac{2}{3} ;-1$

Thus, the correct answer is (b)

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