MCQ
if $x^2+ k(4x + k - 1) + 2 = 0$ has equal $r$ rots, then $k =$
- A$-\frac{2}{3},1$
- ✓$\frac{2}{3},-1$
- C$\frac{3}{2},\frac{1}{3}$
- D$-\frac{3}{2},-\frac{1}{3}$
✓
Answer
Correct option: B.
$\frac{2}{3},-1$
The given quadric equation is $x^2+ k(4x + k - 1) + 2 = 0,$ and roots are equal
Then find the value of $k$.
$x^2+ k(4x + k - 1) + 2 = 0$
$x^2+ 4kx + (k^2- k + 2) = 0$
Here $, a = 1, b = 4k$ and $c = k^2- k + 2$
As we know that $D = b^2- 4ac$
Putting the value of $a = 1, b = 4k$ 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 $D = 0$
$4\left(3 k^2+k-2\right)=0$
$3 k^2+k-2=0$
$3 k^2+3 k-2 k-2=0$
$3k(k + 1) - 2(k + 1) = 0$
$(k + 1)(3k - 2) = 0$
$(k + 1) = 0$ or $(3k - 2) = 0$
$k = -1$ or $\text{k}=\frac{2}{3}$
Therefore, the value of $\text{k}=\frac{2}{3},-1$
Thus, the correct answer is $(b)$
Then find the value of $k$.
$x^2+ k(4x + k - 1) + 2 = 0$
$x^2+ 4kx + (k^2- k + 2) = 0$
Here $, a = 1, b = 4k$ and $c = k^2- k + 2$
As we know that $D = b^2- 4ac$
Putting the value of $a = 1, b = 4k$ 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 $D = 0$
$4\left(3 k^2+k-2\right)=0$
$3 k^2+k-2=0$
$3 k^2+3 k-2 k-2=0$
$3k(k + 1) - 2(k + 1) = 0$
$(k + 1)(3k - 2) = 0$
$(k + 1) = 0$ or $(3k - 2) = 0$
$k = -1$ or $\text{k}=\frac{2}{3}$
Therefore, the value of $\text{k}=\frac{2}{3},-1$
Thus, the correct answer is $(b)$
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