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If $5,5r,5r^{2}$ are the length of the sides of a triangle, then r cannot be equal to:
Option 1)

3/4
Option 2)

5/4
Option 3)

7/4
Option 4)

3/2

Answers (1)

best_answer

Roots of Quadratic Equation -

$\alpha = \frac{-b+\sqrt{b^{2}-4ac}}{2a}$

$\beta = \frac{-b-\sqrt{b^{2}-4ac}}{2a}$

- wherein

$ax^{2}+bx+c= 0$

is the equation

$a,b,c\in R,\: \: a\neq 0$

Quadratic Expression ax^2 + bx + c is non negative -

$ax^{2}+bx+c\geqslant 0$ for all $x \epsilon R$ When $a> 0$ & $b^{2}-4ac\leq 0$ $\left ( \; a,b,c\; \epsilon\; R \right )$

-

5, 5r and 5r² are side of $\Delta 1e$ .

Sum of two side of a triangle is always greater then the third (other) side.

So, when , 0<r<1

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

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

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

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

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

When , r>1

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

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

Option 1)

3/4

Option 2)

5/4

Option 3)

7/4
Option 4)

3/2

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