Answer

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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.

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.

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