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# Solve the quadratic equation $2{{x}^{2}}+x-4=0$ by completing the square.

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Hint: First of all divide the whole equation by 2, to make the coefficients of ${{x}^{2}}=1$. Now to complete the square, try to convert ${{x}^{2}}+\dfrac{x}{2}$ in the form of ${{a}^{2}}+{{b}^{2}}+2ab$ by adding a suitable constant to it and then solve for x.

Complete step by step solution:
Here, we have to solve the quadratic equation $2{{x}^{2}}+x-4=0$ by completing the square.
Let us consider the quadratic equation given in the question.
$2{{x}^{2}}+x-4=0$
First of all, let us divide the whole equation by 2, to get the coefficients of ${{x}^{2}}=1$. We get,
$\dfrac{2{{x}^{2}}}{2}+\dfrac{x}{2}-\dfrac{4}{2}=0$
Or, ${{x}^{2}}+\dfrac{x}{2}-2=0$
We can also write the above equation as,
${{x}^{2}}+2\left( \dfrac{1}{4} \right)\left( x \right)-2=0$
By adding $\dfrac{1}{16}$ to both sides of the above equation, we get,
${{x}^{2}}+2\left( \dfrac{1}{4} \right)x+\dfrac{1}{16}-2=\dfrac{1}{16}$
We can also write $\dfrac{1}{16}={{\left( \dfrac{1}{4} \right)}^{2}}$ in LHS of the above equation, so we get,
$\Rightarrow {{x}^{2}}+2\left( \dfrac{1}{4} \right)x+{{\left( \dfrac{1}{4} \right)}^{2}}-2=\dfrac{1}{16}$
By adding 2 on both sides of the above equation, we get,
$\Rightarrow {{x}^{2}}+2\left( \dfrac{1}{4} \right)x+{{\left( \dfrac{1}{4} \right)}^{2}}=2+\dfrac{1}{16}$
Now, we know that $\left( {{a}^{2}}+2ab+{{b}^{2}} \right)={{\left( a+b \right)}^{2}}$.

So by using this in the above equation and considering a = x and $b=\dfrac{1}{4}$. We get,
$\Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}=\dfrac{33}{16}$
By taking square root on both sides of the above equation, we get,
$\sqrt{{{\left( x+\dfrac{1}{4} \right)}^{2}}}=\pm \sqrt{\dfrac{33}{16}}$
We know that $\sqrt{{{a}^{2}}}=a$. So, we get,
$\left( x+\dfrac{1}{4} \right)=\pm \sqrt{\dfrac{33}{16}}$
By subtracting $\dfrac{1}{4}$ from both sides of the above equation, we get,
$x=\pm \sqrt{\dfrac{33}{16}}-\dfrac{1}{4}$
We know that $\sqrt{16}=4$. So by substituting the value of $\sqrt{16}$ in the above equation, we get,
$x=\pm \dfrac{\sqrt{33}}{4}-\dfrac{1}{4}$
So, we get $x=\dfrac{\sqrt{33}-1}{4}$ and $x=\dfrac{-\sqrt{33}-1}{4}$
Hence, we have solved the quadratic equation $2{{x}^{2}}+x-4$ by completing the square.

Note: Here many students solve the equation by using the quadratic formula that is $x=\dfrac{-b\pm\sqrt{{{b}^{2}}-4ac}}{2a}$ but they must keep in mind that they have to solve the question by completing the square. Though they can cross-check the values of x by using the above formula. Also, it is advisable to convert the coefficient of ${{x}^{2}}=1$ to easily solve the equation by this method.