Question

Find the values of $h$ for which the following equation has real and equal roots:
$x^2+2(h+2)x+9h=0$.

Answer 1

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

$a=1,b=2(h+2)$ and $c=9h$ [0.5 Marks]

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

$D=b^2–4ac$ [0.5 Marks]

$\Rightarrow D=4(h+2{)}^2–4(9h)$
The roots of quadratic equation are real and equal only when the discriminant, $D=0.$ [0.5 Marks]

$\Rightarrow 4(h+2{)}^2–4(9h)=0$

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

Hence, for $h=4$ and $1$, the given equation will have real and equal roots. [0.5 Marks]