Question

5x^2-8x-4=0 when x belongs to Q. solve the quadratic equation

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO SOLVE

The Quadratic equation

$\sf{5 {x}^{2} - 8x - 4 = 0 \: \: \: \: when \: \: x \in \: \mathbb{Q} }$

EVALUATION

Here

$\sf{5 {x}^{2} - 8x - 4 = 0}$

$\implies \sf{5 {x}^{2} - 10x + 2x - 4 = 0}$

$\implies \sf{5x( x - 2) + 2(x - 2) = 0}$

$\implies \sf{( x - 2) (5x + 2) = 0}$

Which gives

$\sf{( x - 2) = 0 \: \: \: or \: \: \: (5x + 2) = 0}$

Now

$\sf{( x - 2) = 0 \: \: gives \: \: x = 2}$

Also

$\displaystyle\sf{( 5x + 2) = 0 \: \: \: gives \: \: x = - \frac{2}{5} }$

Hence the required solution is

$\displaystyle \sf{ \: }x =- \frac{2}{5} \: , \: 2 \: \: \: \: \: \: when \: \: \: \: x \in \: \mathbb{Q}$

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Answer 2

The given quadratic equation:

5$x^2$ - 8x - 4 = 0

Here, a = 5, b = - 8 and c = - 4

We have to solve the given quadratic equation.

Solution:

∴ Discriminant, D = $b^{2} -4ac$

= $(-8)^2-4(5)(-4)$

= 64 + 80

= 144 > 0, the two roots are real and unequal.

∴ $x=\dfrac{-b±\sqrt{D} }{2a}$

     $=\dfrac{-(-8)±\sqrt{144} }{2(5)}$

     $=\dfrac{8±12 }{10}$

    $=\dfrac{8+12 }{10}, \dfrac{8-12 }{10}$

   $=\dfrac{20 }{10}, \dfrac{-4 }{10}$

x = 2 or, $-\dfrac{2}{5}$

Thus, x =  x = 2 or, $-\dfrac{2}{5}$