Solve the quadratic equation \[2{{x}^{2}}+x-4=0\] by completing the square.
Last updated date: 26th Mar 2023
•
Total views: 306.6k
•
Views today: 4.83k
Answer
306.6k+ views
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.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
