Question

If the equation $x^2+2(k+2)x+9k=0$ has equal real roots, then value(s) of k are __________.

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (A), (D)

The correct options are
A 4
D 1

For, $x^2+2(k+2)x+9k=0$,
a = 1, b = 2(k +2) and c = 9k.

$D=b^2-4ac=\left[2\right(k+2){]}^2–4(1\left)\right(9k)=4(k+2{)}^2–4\left(9k\right)=4(k^2+4+4k-9k)=4(k^2+4-5k)$

The roots of quadratic equation are real and equal only when $D=0$.

$\Rightarrow 4(k^2+4-5k)=0$
$\Rightarrow k^2-5k+4=0$

On comparing with
$a{x}^2+bx+c=0$, we get
a= 1, b = -5 and c = 4 and x = k

By quadratic formula,
$k=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$k=\frac{-(-5)\pm \sqrt{(-5{)}^2-4(1\left)\right(4)}}{2\times 1}$

$k=\frac{5\pm \sqrt{25-16}}{2}$

$k=\frac{5\pm \sqrt{9}}{2}$

$k=\frac{5\pm 3}{2}$

$k=\frac{5+3}{2}$ or $k=\frac{5-3}{2}$

​​​​​​​$k=\frac{8}{2}$ or ​​​​​​​$k=\frac{2}{2}$

$k=4$ or $k=1$