Mathematics - Quadratic Equation Question with Solution | TestHub

Let $b=ar$ and $c=a{r}^2$, since $a$, $b$, $c$ are in a Geometric Progression (G.P.).

Therefore, we have:

$$a+b+c=xb$$

$$a+ar+a{r}^2=x\cdot ar$$

Dividing by $a$ (assuming $a\ne 0$), we get:

$$1+r+r^2=xr$$

Rearranging the terms, we obtain a quadratic equation in $r$:

$$r^2+(1-x)r+1=0$$

Since $r$ is a real number, the discriminant of this quadratic equation must be greater than or equal to zero.

$$Disc.\ge 0$$

$$(1-x{)}^2-4\cdot 1\cdot 1\ge 0$$

$$(1-x{)}^2-4\ge 0$$

Expanding and simplifying:

$$1-2x+x^2-4\ge 0$$

$$x^2-2x-3\ge 0$$

Factoring the quadratic expression:

$$(x+1)(x-3)\ge 0$$

This inequality holds true when $x\le -1$ or $x\ge 3$.