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(C) For $5, 5r, 5r^2$ to be the sides of a triangle,the sum of any two sides must be greater than the third side.$1) 5 + 5r > 5r^2 \Rightarrow r^2 - r

Vedclass · Accepted Answer

$\frac{7}{4}$

Vedclass · Answer

(C) For $5, 5r, 5r^2$ to be the sides of a triangle,the sum of any two sides must be greater than the third side.$1) 5 + 5r > 5r^2 \Rightarrow r^2 - r - 1  0$,we have $0 $2) 5 + 5r^2 > 5r \Rightarrow r^2 - r + 1 > 0$. This is true for all $r \in \mathbb{R}$ as the discriminant $D = (-1)^2 - 4(1)(1) = -3 $3) 5r + 5r^2 > 5 \Rightarrow r^2 + r - 1 > 0$. The roots of $r^2 + r - 1 = 0$ are $r = \frac{-1 \pm \sqrt{5}}{2}$. Since $r > 0$,we have $r > \frac{-1 + \sqrt{5}}{2} \approx 0.618$.Combining these,we get $\frac{\sqrt{5} - 1}{2} Checking the options:$A) \frac{3}{4} = 0.75$ (in range)$B) \frac{5}{4} = 1.25$ (in range)$C) \frac{7}{4} = 1.75$ (outside range)$D) \frac{3}{2} = 1.5$ (in range)Thus,$r$ cannot be $\frac{7}{4}$.

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

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