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Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
$2x^2 +x + 4 = 0$
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 $(a+b)^2=a^2+2ab+b^2$)
$(x+\frac{1}{4})^2=\frac{1-2\times16}{16}$
$(x+\frac{1}{4})^2=\frac{1-32}{16}$
$(x+\frac{1}{4})^2=\frac{-31}{16}$
$x+\frac{1}{4}=\pm \sqrt{\frac{-31}{16}}$ (Taking square root on both sides)
$x=\sqrt{\frac{-31}{16}}-\frac{1}{4}$ or $x=-\sqrt{\frac{-31}{16}}-\frac{1}{4}$
$x=\frac{\sqrt{-31}-1}{4}$ or $x=-(\frac{\sqrt{-31}+1}{4})$
Therefore, no real roots exist for the given quadratic equation.