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)
A admin

 

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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