# Find the roots of the following quadratic equations, if they exist, by the method of completing the square:$2x^2 + x - 4 = 0$

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Given:

Given quadratic equation is $2x^2+x -4 = 0$.

To do:

We have to find the roots of the given quadratic equation.

Solution:

$2x^2+x - 4 = 0$

$2(x^2 + \frac{1}{2} x -\frac{4}{2}) = 0$

$x^2 + \frac{1}{2} x -2 = 0$

$x^2 + 2\times \frac{1}{2}\times \frac{1}{2} x = 2$

$x^2 + 2\times \frac{1}{4} x = 2$

Adding $(\frac{1}{4})^2$ on both sides completes the square. Therefore,

$x^2 + 2\times (\frac{1}{4}) x + (\frac{1}{4})^2 = 2+(\frac{1}{4})^2$

$(x+\frac{1}{4})^2=2+\frac{1}{16}$      (Since $(a+b)^2=a^2+2ab+b^2$)

$(x+\frac{1}{4})^2=\frac{1+2\times16}{16}$

$(x+\frac{1}{4})^2=\frac{1+32}{16}$

$(x+\frac{1}{4})^2=\frac{33}{16}$

$x+\frac{1}{4}=\pm \sqrt{\frac{33}{16}}$     (Taking square root on both sides)

$x=\sqrt{\frac{33}{16}}-\frac{1}{4}$ or $x=-\sqrt{\frac{33}{16}}-\frac{1}{4}$

$x=\frac{\sqrt{33}-1}{4}$ or $x=-(\frac{\sqrt{33}+1}{4})$

The values of $x$ are $\frac{\sqrt{33}-1}{4}$ and $-(\frac{\sqrt{33}+1}{4})$.

Updated on 10-Oct-2022 13:20:12