Question

How do you solve \[0=3{{x}^{2}}-2x-12\] by completing the square?

Answer 1

Hint: This question is from the topic of quadratic equation. In this question, we have to find the value of x. In solving this question, we will first add and subtract a term to the right side of the equation, so that we can make the equation in the form of \[{{a}^{2}}-2\times a\times b+{{b}^{2}}\]. From there, we will make an equation in the form of \[{{\left( a-b \right)}^{2}}\]. From there, we will get the value of x.

Complete step by step solution:
Let us solve this question.
In this question, we have to solve the equation \[0=3{{x}^{2}}-2x-12\] by completing the square. We will find the value of x from the given equation.
The given equation is
\[0=3{{x}^{2}}-2x-12\]
Let us make the right side of equation in the form of \[{{a}^{2}}-2\times a\times b+{{b}^{2}}\] to make a perfect square. We can write the above equation as
\[\Rightarrow 0={{\left( \sqrt{3}x \right)}^{2}}-2\times \left( \sqrt{3}x \right)\times \left( \dfrac{1}{\sqrt{3}} \right)+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}-12\]
Here, we can see that we have multiplied and divided a term that is \[\sqrt{3}\] with the term ‘2x’ and added & subtracted the term which is square of \[\dfrac{1}{\sqrt{3}}\] to make a complete square.
Now, we can see in the above equation that first three terms of the right side of the equation are in the form of \[{{a}^{2}}-2\times a\times b+{{b}^{2}}\], and we know that we can write the equation \[{{a}^{2}}-2\times a\times b+{{b}^{2}}\] as \[{{\left( a-b \right)}^{2}}\].
So, we can write the above equation as
\[\Rightarrow 0={{\left( \sqrt{3}x-\dfrac{1}{\sqrt{3}} \right)}^{2}}-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}-12\]
The above equation can also be written as
\[\Rightarrow 0={{\left( \sqrt{3}x-\dfrac{1}{\sqrt{3}} \right)}^{2}}-\dfrac{1}{3}-12\]
The above equation can also be written as
\[\Rightarrow 0={{\left( \sqrt{3}x-\dfrac{1}{\sqrt{3}} \right)}^{2}}-\dfrac{37}{3}\]
The above equation can also be written as
\[\Rightarrow \dfrac{37}{3}={{\left( \sqrt{3}x-\dfrac{1}{\sqrt{3}} \right)}^{2}}\]
The above equation can also be written as
\[\Rightarrow {{\left( \sqrt{3}x-\dfrac{1}{\sqrt{3}} \right)}^{2}}=\dfrac{37}{3}\]
Taking square root to the both side of the equation, we get
\[\Rightarrow \left( \sqrt{3}x-\dfrac{1}{\sqrt{3}} \right)=\pm \sqrt{\dfrac{37}{3}}=\pm \dfrac{\sqrt{37}}{\sqrt{3}}\]
Now, we can write the above equation as
\[\Rightarrow \sqrt{3}x=\dfrac{1}{\sqrt{3}}\pm \dfrac{\sqrt{37}}{\sqrt{3}}\]
Now, dividing \[\sqrt{3}\] to the both side of equation, we get
\[\Rightarrow x=\dfrac{1}{\sqrt{3}\times \sqrt{3}}\pm \dfrac{\sqrt{37}}{\sqrt{3}\times \sqrt{3}}\]
The above equation can also be written as
\[\Rightarrow x=\dfrac{1}{3}\pm \dfrac{\sqrt{37}}{3}=\dfrac{1\pm \sqrt{37}}{3}\]
Hence, the values of x are \[\dfrac{1+\sqrt{37}}{3}\] and \[\dfrac{1-\sqrt{37}}{3}\].

Now, we have solved the equation \[0=3{{x}^{2}}-2x-12\] and found the value of x as \[\dfrac{1+\sqrt{37}}{3}\] and \[\dfrac{1-\sqrt{37}}{3}\].

Note: We should have a better knowledge in the topic of quadratic equations to solve this type of question easily. We should know how to solve a quadratic equation by completing the square. Always remember the following formula:
\[{{a}^{2}}-2\times a\times b+{{b}^{2}}={{\left( a-b \right)}^{2}}\]
If we want to check if we are correct or not, then using the Sridharacharya method we can check that.
So, according to Sridharacharya’s method the value of x for the equation \[a{{x}^{2}}+bx+c=0\] is in the following:
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
The equation \[0=3{{x}^{2}}-2x-12\] can also be written as \[3{{x}^{2}}-2x-12=0\]
So, the value of x for the equation \[3{{x}^{2}}-2x-12=0\] is
\[x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 3\times \left( -12 \right)}}{2\times 3}=\dfrac{2\pm \sqrt{4+4\times 36}}{2\times 3}\]
The above can also be written as
\[x=\dfrac{2\pm 2\sqrt{1+36}}{2\times 3}=\dfrac{1\pm \sqrt{1+36}}{3}=\dfrac{1\pm \sqrt{37}}{3}\]
Hence, we get that the values of x are \[\dfrac{1+\sqrt{37}}{3}\] and \[\dfrac{1-\sqrt{37}}{3}\]
So, our answer is correct.