Question

Solve the equation $\mathrm{kx}(\mathrm{x}-2)+6=0$ for the value of $\mathrm{x}$ if roots of this equation are real and equal.

Answer 1

Given equation is : $\mathrm{kx}(\mathrm{x}-2)+6=0$

⇒ ${\mathrm{kx}}^2-2\mathrm{kx}+6=0$

Follow these steps of the solution:

Step $1$:

On comparing the given equation with the general form of the equation ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$, we get

$\mathrm{a}=\mathrm{k}$, $\mathrm{b}=-2\mathrm{k}$,$\mathrm{c}=6$

We know that for the real and equal roots of any quadratic equation(Discriminant) $\mathrm{D}=0$

Then ${\mathrm{b}}^2-4\mathrm{ac}=0$ (Formula used: $\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac}$)

${(-2\mathrm{k})}^2-4\mathrm{k}\left(6\right)=0$

$4{\mathrm{k}}^2-24\mathrm{k}=0$

$4\mathrm{k}(\mathrm{k}-6)=0$

$\Rightarrow 4\mathrm{k}=0$ or $\mathrm{k}=0$

Also$(\mathrm{k}-6)=0$ or $\mathrm{k}=6$

Step $2$:

Neglecting the value $\mathrm{k}=0$ as it will not form any quadratic equation.

Now put the value of $\mathrm{k}$in equation (i) we get

$6{\mathrm{x}}^2-12\mathrm{x}+6=0$

$6{\mathrm{x}}^2-6\mathrm{x}-6\mathrm{x}+6=0$ (Factorizing by splitting the middle term)

$\Rightarrow 6\mathrm{x}(\mathrm{x}-1)-6(\mathrm{x}-1)=0$

$\Rightarrow (6\mathrm{x}-6)(\mathrm{x}-1)$ $=0$

or $6\mathrm{x}-6$ $=0$

$6\mathrm{x}$ $=6$

$\mathrm{x}$ $=\frac{6}{6}$

$\Rightarrow$$\mathrm{x}$ $=1$

Similarly,$(\mathrm{x}-1)$ $=0$

$\Rightarrow \mathrm{x}$ $=1$

Thus, $\mathrm{x}$$=1$ is the required solution