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) $\frac{7}{4}$

(c) $\frac{3}{2}$

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

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Answer:

Correct Answer: 9. (b)

Solution:

  1. 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.

$\therefore a+b>c ; b+c>a$ and $c+a>b$

Now, $\quad a+b>c$

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

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

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

$\left[\because\right.$ 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} \text { and } r^{2}-r-1=0 \\ \left.\Rightarrow r=\frac{1 \pm \sqrt{1+4}}{2}=\frac{1 \pm \sqrt{5}}{2}\right] \end{array} $

$ \begin{aligned} \Rightarrow \quad 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, $\quad b+c>a$

$ \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 $

$ \frac{-1-\sqrt{5}}{2} \frac{-1+\sqrt{5}}{2}$

$ \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 $

$\Rightarrow \quad r \in R$

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

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

and $\frac{7}{4}$ is the only value that does not satisfy.



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