Question

Solve the quadratic equation \[9{{x}^{2}}-15x+6=0\] by the method of completing the square.
(a) \[x=-1\text{ }and\text{ }x=\dfrac{2}{3}\]
(b) \[x=1\text{ }and\text{ }x=\dfrac{1}{3}\]
(c) \[x=1\text{ }and\text{ }x=\dfrac{2}{3}\]
(d) None of these

Answer 1

Hint: First of all divide the whole equation by 9 to make the coefficient of \[{{x}^{2}}\] equal to 1. Now, try to compute the square by converting \[{{x}^{2}}=\dfrac{15x}{9}\] in the form of \[{{a}^{2}}+{{b}^{2}}+2ab\] by adding a suitable constant to it and then solve the value for x.

Complete step-by-step answer:
Here, we have to solve the quadratic equation \[9{{x}^{2}}-15x+6=0\] by the method of completing the square. Let us consider the quadratic equation given in the question.
\[9{{x}^{2}}-15x+6=0\]
First of all, let us divide the whole equation by 9 to get the coefficient of \[{{x}^{2}}\] equal to 1, we get,
\[\dfrac{9{{x}^{2}}-15x+6}{9}=0\]
\[\Rightarrow {{x}^{2}}-\dfrac{15x}{9}+\dfrac{6}{9}=0\]
\[\Rightarrow {{x}^{2}}-\dfrac{5}{3}x+\dfrac{2}{3}=0\]
We can also write the above equation as:
\[\Rightarrow {{x}^{2}}-2x\left( \dfrac{5}{6} \right)+\dfrac{2}{3}=0\]
By adding \[{{\left( \dfrac{5}{6} \right)}^{2}}\] on both sides of the above equation, we get,
\[{{x}^{2}}-2\left( \dfrac{5}{6} \right)x+\dfrac{2}{3}+{{\left( \dfrac{5}{6} \right)}^{2}}={{\left( \dfrac{5}{6} \right)}^{2}}\]
By subtracting \[\dfrac{2}{3}\] on both the sides of the above equation, we get,
\[{{x}^{2}}-2\left( \dfrac{5}{6} \right)x+{{\left( \dfrac{5}{6} \right)}^{2}}=\dfrac{25}{36}-\dfrac{2}{3}\]
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 \[b=\dfrac{5}{6}\], we get,
\[{{\left( x-\dfrac{5}{6} \right)}^{2}}=\dfrac{25-24}{36}\]
\[{{\left( x-\dfrac{5}{6} \right)}^{2}}=\dfrac{1}{36}\]
By taking the square root on both the sides of the above equation, we get,
\[\sqrt{{{\left( x-\dfrac{5}{6} \right)}^{2}}}=\pm \sqrt{\dfrac{1}{36}}\]
We know that \[\sqrt{{{a}^{2}}}=a\]. So, we get,
\[\left( x-\dfrac{5}{6} \right)=\pm \dfrac{1}{6}\]
By adding \[\dfrac{5}{6}\] on both the sides of the above equation, we get,
\[x=\pm \dfrac{1}{6}+\dfrac{5}{6}\]
So, we get, \[x=\dfrac{1}{6}+\dfrac{5}{6}=\dfrac{6}{6}=1\] and \[x=\dfrac{-1}{6}+\dfrac{5}{6}=\dfrac{4}{6}=\dfrac{2}{3}\]
Hence, we have solved the quadratic equation \[9{{x}^{2}}-15x+6=0\] and got \[x=1,\dfrac{2}{3}\]
Therefore, option (c) is the right answer.

Note: Here, many students solve the quadratic 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 their values of x by using the above formula, it is advisable to convert the coefficient of \[{{x}^{2}}=1\] to easily solve the question by this method.