SATHEE: Theory of Equations 1 Question 8

Theory of Equations 1 Question 8

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

(a) $\frac{5}{4}$

(b) $\frac{7}{4}$

(d) $\frac{3}{4}$

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Solution:

Let $a = 5, b = 5 r$ and $c = 5 r^{2}$

We know that, in a triangle sum of 2 sides is always greater than the third side.

$∴ a + b > c; b + c > a$ and $c + a > b$

Now, $a + b > c$

$\Rightarrow 5 + 5 r > 5 r^{2} \Rightarrow 5 r^{2} - 5 r - 5 < 0$

$\Rightarrow r^{2} - r - 1 < 0$

$\Rightarrow r - \frac{1 - \sqrt{5}}{2} r - \frac{1 + \sqrt{5}}{2} < 0$

$[∵$ roots of $a x^{2} + b x + c = 0$ are given by

$\begin{array}{r} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} and r^{2} - r - 1 = 0 \\ \Rightarrow r = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}] \end{array}$

$\begin{aligned} \Rightarrow r & \in \frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2} \\ \frac{1}{\frac{1 - \sqrt{5}}{2}} - \frac{1 + \sqrt{5}}{2} \end{aligned}$

Similarly, $b + c > a$

$$ \begin{aligned} & \Rightarrow \quad 5 r+5 r^{2}>5 \ & \Rightarrow \quad r^{2}+r-1>0 \ & r-\frac{-1-\sqrt{5}}{2} \quad r-\frac{-1+\sqrt{5}}{2}>0 \ & \because r^{2}+r-1=0 \Rightarrow r=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2} \ & \Rightarrow \quad r \in-\infty, \frac{-1-\sqrt{5}}{2} \cup \frac{-1+\sqrt{5}}{2}, \infty \ & \begin{array}{ccc}

& - & + \ \hline \frac{-1-\sqrt{5}}{2} & & \frac{-1+\sqrt{5}}{2} \end{array} \ & \text { and } \quad c+a>b \ & \Rightarrow \quad 5 r^{2}+5>5 r \ & \Rightarrow \quad r^{2}-r+1>0 \ & \Rightarrow r^{2}-2 \cdot \frac{1}{2} r+\frac{1}{2}^{2}+1-\frac{1}{2}^{2}>0 \ & \Rightarrow \quad r-\frac{1}{2}^{2}+\frac{3}{4}>0 \ & \Rightarrow \quad r \in R \end{aligned} $$

From Eqs. (i), (ii) and (iii), we get

$\begin{aligned} r \in \frac{- 1 + \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2} \\ \underset{- \infty}{⟵} \end{aligned}$