Question

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

• A. $1and4$
• B. $–1and5$
• C. $1and–5$

Answer 1

The correct option is A $1and4$

On comparing $x^2+2(k+2)x+9k=0$ with standard form $a{x}^2+bx+c=0$, we get

a = 1, b = 2(k + 2) and c = 9k

For, $x^2+2(k+2)x+9k=0$, value of discriminant,

$D=b^2–4ac$

$\Rightarrow D=4(k+2{)}^2–4(9k)$
The roots of quadratic equation are real and equal only when discriminant, $D=0.$

$\Rightarrow 4(k+2{)}^2–4(9k)=0$

$\Rightarrow (k+2{)}^2=9k\Rightarrow k^2-5k+4=0\Rightarrow k^2-4k-k+4=0\Rightarrow k^2-k-4k+4=0\Rightarrow k(k-1)-4(k-1)=0\Rightarrow (k-4\left)\right(k-1)=0\therefore k=4or1$