Question

How do you solve \[0=3{{x}^{2}}-11x+6\] by completing the square?

Answer 1

Hint: In algebra, completing square is a method to convert a quadratic polynomial of the form\[a{{x}^{2}}+bx+c\] to the form \[a{{\left( x-h \right)}^{2}}+k\] where \[h,k\]are constants. Here, we will apply the basic property that\[\sqrt{{{x}^{2}}}=x\].

Complete step by step solution:
From the question it is clear that we have to solve \[0=3{{x}^{2}}-11x+6\] by completing the square method.
In algebra, completing square is a method to convert a quadratic polynomial of the form\[a{{x}^{2}}+bx+c\] to the form \[a{{\left( x-h \right)}^{2}}+k\] where \[h,k\]are constants.
\[0=3{{x}^{2}}-11x+6\] it is clearly in the form of \[a{{x}^{2}}+bx+c\]. Now we will try to reduce the given equation into \[a{{\left( x-h \right)}^{2}}+k\].
Consider the given equation
\[\Rightarrow 0=3{{x}^{2}}-11x+6\]…………….(1)
Now take 3 common on RHS part.
\[\Rightarrow 0=3\left( {{x}^{2}}-\dfrac{11}{3}x+\dfrac{6}{3} \right)\]
Now divide with 3 on both sides
\[\Rightarrow \dfrac{0}{3}=\dfrac{3}{3}\left( {{x}^{2}}-\dfrac{11}{3}x+\dfrac{6}{3} \right)\]
After simplification we get
\[\Rightarrow 0={{x}^{2}}-\dfrac{11}{3}x+2\]……………..(2)
Now add and subtract \[\dfrac{121}{36}\]on the RHS part. Now we will get
\[\Rightarrow 0={{x}^{2}}-\dfrac{11}{3}x+\dfrac{121}{36}-\dfrac{121}{36}+2\]
Now consider the term \[\dfrac{11}{3}x\]. Multiply and divide with 2. So \[\dfrac{11}{3}x\] is converted into \[\dfrac{2}{2}\times \dfrac{11}{3}x\]
Further it can be written as \[2\times \dfrac{11}{6}x\]
Now put \[\dfrac{11}{3}x=2\times \dfrac{11}{6}x\].
So, the equation becomes
\[\Rightarrow 0={{x}^{2}}-2\times \dfrac{11}{6}x+\dfrac{121}{36}-\dfrac{121}{36}+2\]
Now subtract \[\dfrac{121}{36}\] from \[2\]. We will get \[-\dfrac{49}{36}\]
\[\Rightarrow 0={{x}^{2}}-\dfrac{11}{3}x+\dfrac{121}{36}-\dfrac{49}{36}\]………………..(3)
We know that \[121={{\left( 11 \right)}^{2}}\]and \[36={{\left( 6 \right)}^{2}}\].
So \[\dfrac{121}{36}\] can be written as \[\dfrac{{{\left( 11 \right)}^{2}}}{{{\left( 6 \right)}^{2}}}\].
Now from the basic concepts of mathematics we can write \[\dfrac{{{\left( a \right)}^{2}}}{{{\left( b \right)}^{2}}}={{\left( \dfrac{a}{b} \right)}^{2}}\].
Now we can write \[\dfrac{{{\left( 11 \right)}^{2}}}{{{\left( 6 \right)}^{2}}}={{\left( \dfrac{11}{6} \right)}^{2}}\]
So, the equation becomes
\[\Rightarrow 0={{x}^{2}}-2\times \dfrac{11}{6}x+{{\left( \dfrac{11}{6} \right)}^{2}}-\dfrac{49}{36}\]…………………..(4)
Observe the part \[{{x}^{2}}-2\times \dfrac{11}{6}x+{{\left( \dfrac{11}{6} \right)}^{2}}\] clearly. We can it is in the form of \[{{a}^{2}}-2ab+{{b}^{2}}\]
From the basic algebraic formula, we know \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\].
So \[{{x}^{2}}-2\times \dfrac{11}{6}x+{{\left( \dfrac{11}{6} \right)}^{2}}\] can be written in the form of \[{{\left( a-b \right)}^{2}}\]
Here \[a=x\] and \[b=\dfrac{11}{6}\]
So \[{{x}^{2}}-2\times \dfrac{11}{6}x+{{\left( \dfrac{11}{6} \right)}^{2}}\] can be written as \[{{\left( x-\dfrac{11}{6} \right)}^{2}}\]
Put\[{{x}^{2}}-2\times \dfrac{11}{6}x+{{\left( \dfrac{11}{6} \right)}^{2}}={{\left( x-\dfrac{11}{6} \right)}^{2}}\] in the equation (4)
\[\Rightarrow 0={{\left( x-\dfrac{11}{6} \right)}^{2}}-\dfrac{49}{36}\]
Now add \[\dfrac{49}{36}\]on both sides, we get
\[\Rightarrow 0+\dfrac{49}{36}={{\left( x-\dfrac{11}{6} \right)}^{2}}-\dfrac{49}{36}+\dfrac{49}{36}\]
After simplification we, get
\[\Rightarrow \dfrac{49}{36}={{\left( x-\dfrac{11}{6} \right)}^{2}}\]
Now apply square root on both sides
\[\Rightarrow \sqrt{\dfrac{49}{36}}=\sqrt{{{\left( x-\dfrac{11}{6} \right)}^{2}}}\]
we will apply the basic property that\[\sqrt{{{x}^{2}}}=x\].
\[\Rightarrow \pm \dfrac{7}{6}=x-\dfrac{11}{6}\]
\[\Rightarrow x=\dfrac{11}{6}\pm \dfrac{7}{6}\]
We will get two \[x\] values
\[x=\dfrac{11}{6}+\dfrac{7}{6}\] or \[x=\dfrac{11}{6}-\dfrac{7}{6}\]
\[x=\dfrac{11+7}{6}\] or \[x=\dfrac{11-7}{6}\]
\[x=\dfrac{18}{6}\] or \[x=\dfrac{4}{6}\]
After simplification we get
\[x=3\] or \[x=\dfrac{2}{3}\]
Hence the question is solved.

NOTE:
While solving this kind of question we have to know the exact values of square root. And we need to know about some basic concepts of mathematics. students have to be careful while doing calculations.
Even small calculation mistakes can lead to large errors in the final answer.