# Solve this following

Question:

If $5,5 \mathrm{r}, 5 \mathrm{r}^{2}$ are the lengths of the sides of a triangle, then $\mathrm{r}$ cannot be equal to:

1. $\frac{3}{2}$

2. $\frac{3}{4}$

3. $\frac{5}{4}$

4. $\frac{7}{4}$

Correct Option: , 4

Solution:

$\mathrm{r}=1$ is obviously true.

Let $0<\mathrm{r}<1$

$\Rightarrow \quad r+r^{2}>1$

$\Rightarrow r^{2}+r-1>0$

$\left(r-\frac{-1-\sqrt{5}}{2}\right)\left(r-\left(\frac{-1+\sqrt{5}}{2}\right)\right)$

$\Rightarrow r-\frac{-1-\sqrt{5}}{2}$ or $r>\frac{-1+\sqrt{5}}{2}$

$r \in\left(\frac{\sqrt{5}-1}{2}, 1\right)$

$\frac{\sqrt{5}-1}{2} When$r>1\Rightarrow \frac{\sqrt{5}+1}{2}>\frac{1}{\mathrm{r}}>1\Rightarrow \mathrm{r} \in\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)\$

Now check options