Question

2x^2-5x-4=0, find root of quadratic equation

Answer 1

Answer:

Answer:

Roots of the given quadratic equation are

$x=\frac{5+\sqrt{57} }{4}\ or\ x=\frac{5-\sqrt{57} }{4}$.

Step-by-step explanation:

We find the roots by using quadratic formula  

$x=\frac{-b}{2a}$±$\frac{\sqrt{b^{2}-4ac} }{2a}$,

in the given equation $2x^{2} -5x-4=0$,

the value of $a=2,b=-5,c=-4$.

Substituting in the formula we get

$x=\frac{5}{4}$±$\frac{\sqrt{(-5)^{2}-4(2)(-4)}}{4}$

  $=\frac{5}{4}$±$\frac{\sqrt{25+32} }{4}$

  $=\frac{5}{4}$±$\frac{\sqrt{57} }{4}$.

Thus the values of $x$ are $x=\frac{5}{4} +\frac{\sqrt{57} }{4}$ or $x=\frac{5-\sqrt{57} }{4}$.

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