Question

How do you solve by completing the square $9{{x}^{2}}+6x=8$?

Answer 1

In the given question, we have been asked to find the value of ‘x’ by solving the given equation i.e. $9{{x}^{2}}+6x=8$ using the completing the square method. Completing the square method is used to solve the quadratic equation by converting the form of the equation so that it will become a perfect trinomial. And then after applying the square formula, we will get our required solution.

Formula used:
${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$

Complete step by step solution:
We have given that,
$\Rightarrow 9{{x}^{2}}+6x=8$
Dividing both the side of the equation by 9, we get
$\Rightarrow {{x}^{2}}+\dfrac{6}{9}=\dfrac{8}{9}$
Simplifying the above, we get
$\Rightarrow {{x}^{2}}+\dfrac{2}{3}=\dfrac{8}{9}$
Now, for completing the square adding ${{\left( \dfrac{1}{3} \right)}^{2}}$ to both the sides of the equation, we get

\right)}^{2}}$$ $$\Rightarrow {{x}^{2}}+\dfrac{2}{3}+{{\left( \dfrac{1}{3} \right)}^{2}}=\dfrac{8}{9}+\dfrac{1}{9}$$ Simplifying the above equation, we get $$\Rightarrow {{x}^{2}}+\dfrac{2}{3}+{{\left( \dfrac{1}{3} \right)}^{2}}=\dfrac{9}{9}$$ $$\Rightarrow {{x}^{2}}+\dfrac{2}{3}+{{\left( \dfrac{1}{3} \right)}^{2}}=1$$ As we know that, $${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$$ Therefore, $$\Rightarrow {{\left( x+\dfrac{1}{3} \right)}^{2}}=1$$ Transposing the power 2 on the right side of the equation, we get $$\Rightarrow \left( x+\dfrac{1}{3} \right)=\pm \sqrt{1}$$ As we know that $$\pm \sqrt{1}=\pm 1$$ Simplifying the above, we get $$\Rightarrow x=-\dfrac{1}{3}\pm 1$$ Now, we have $$\Rightarrow x=-\dfrac{1}{3}+1$$ or $$x=-\dfrac{1}{3}-1$$ Now, solving $$\Rightarrow x=-\dfrac{1}{3}+1$$ Taking LCM of 3 and 1 which is equal to 3, we get $$\Rightarrow x=\dfrac{-1+3}{3}=\dfrac{2}{3}$$ Now, solving $$\Rightarrow x=-\dfrac{1}{3}-1$$ Taking LCM of 3 and 1 which is equal to 3, we get $$\Rightarrow x=\dfrac{-1-3}{3}=\dfrac{-4}{3}$$ $$\Rightarrow x=\dfrac{2}{4}\ or\ \dfrac{-4}{3}$$ **Therefore, the possible values of $$x\ are\ \dfrac{2}{4}\ or\ \dfrac{-4}{3}$$** **Note:** While solving these types of questions, students should carefully observe when considering the terms, which one is the ‘a’ and the ‘b’. To check whether the obtained possible values are correct or not, we can verify the result by solving the quadratic equation with the roots of the quadratic equation formula. Standard form of quadratic equation; $$a{{x}^{2}}+bx+c=0$$, then the roots of the quadratic equation is given by, $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$.