# $Cos\Theta$ is a root of the equation $25x^{2}+5x-12=0,\:\:-1< x< 0,$ then the value of $Sin2\theta$ is, Option 1) $\frac{24}{25}$ Option 2) $\frac{-12}{25}$ Option 3) $\frac{-28}{25}$ Option 4) $\frac{20}{25}$

As we learnt in concept

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

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

- wherein

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

is the equation

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

$25 x^{2}+5x-12 =0$

$cos \theta$ is the root of this equation

Roots of this equations are :

$\frac{-5 \pm \sqrt{25+4*12*25}}{50}$

$=\frac{-5 \pm \sqrt{25*49}}{50}$

$=\frac{-5\pm 35}{50}$

$=\frac{3}{5}, \frac{-4}{5}$

$\cos \theta= \frac{3}{5}\: \: , \sin \theta = \frac{4}{5}$

Thus , $\sin \2\theta= 2 \sin \theta \cos \theta$

$= 2*\frac{3}{5}*\frac{4}{5}$

$=\frac{24}{25}$

Option 1)

$\frac{24}{25}$

Correct option

Option 2)

$\frac{-12}{25}$

Incorrect option

Option 3)

$\frac{-28}{25}$

Incorrect option

Option 4)

$\frac{20}{25}$

Incorrect option

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