Question

How do you solve using the completing the square method $3{{x}^{2}}-7x-6=0$?

Answer 1

In this problem, we have to find the value of x by solving the given quadratic equation in completing the square method. For this, we can first convert the given quadratic equation into a complete square equation by algebraic whole square formula. we can add the missing term on both sides to get a complete square equation on the left hand side, then we will get a whole square form, for which we can take square root on both sides to get the value of x.

Complete step by step solution:
We know that the given quadratic equation to be solved is,
$3{{x}^{2}}-7x-6=0$ ……. (1)
Now we can add 3 on both sides in the above equation (1), we get
$\Rightarrow 3{{x}^{2}}-7x-6+6=6$
Now we can divide on both sides to get a perfect square equation, we get

& \Rightarrow \dfrac{3{{x}^{2}}}{3}-\dfrac{7x}{3}=\dfrac{6}{3} \\\ & \Rightarrow {{x}^{2}}-\dfrac{7}{3}x=2.......(2) \\\ \end{aligned}$$ We can now take the first two terms $${{x}^{2}}-\dfrac{7}{3}x$$ We know that, $${{\left( x-a \right)}^{2}}={{x}^{2}}-2ax+{{a}^{2}}$$ We can see that, $$\begin{aligned} & \Rightarrow -2a=-\dfrac{7}{3} \\\ & \Rightarrow a=\dfrac{7}{6} \\\ & \Rightarrow {{a}^{2}}={{\left( \dfrac{7}{6} \right)}^{2}}=\dfrac{49}{36} \\\ \end{aligned}$$ We can add the above value in both the left-hand side and the right-hand side of equation (2), to get a perfect square equation, we get $$\Rightarrow {{x}^{2}}-\dfrac{7}{3}x+\dfrac{49}{36}=2+\dfrac{49}{36}$$ We can write the left-hand side in whole square form as it is a perfect square equation, we get $$\Rightarrow {{\left( x-\dfrac{7}{6} \right)}^{2}}=\dfrac{2\times 36+49}{36}=\dfrac{121}{36}={{\left( \dfrac{11}{6} \right)}^{2}}$$ Now we can take square root on both sides, we get $$\begin{aligned} & \Rightarrow x-\dfrac{7}{6}=\pm \dfrac{11}{6} \\\ & \Rightarrow x=\dfrac{7}{6}\pm \dfrac{11}{6} \\\ \end{aligned}$$ Now we can simplify the above step, we get $$\begin{aligned} & \Rightarrow x=\dfrac{7+11}{6}=\dfrac{18}{6}=3 \\\ & \Rightarrow x=\dfrac{7-11}{6}=\dfrac{-2}{3} \\\ \end{aligned}$$ **Therefore, the value of $$x=3,-\dfrac{2}{3}$$.** **Note:** We can now verify to check whether the values we got are correct or not. We can substitute the values of x in the equation to check. We can now take the equation (1) and substitute the value of $$x=3,-\dfrac{2}{3}$$, we get When x = 3, $$\begin{aligned} & \Rightarrow 3{{\left( 3 \right)}^{2}}-7\left( 3 \right)-6=0 \\\ & \Rightarrow 27-27=0 \\\ \end{aligned}$$ When $$x=\dfrac{-2}{3}$$, $$\begin{aligned} & \Rightarrow 3{{\left( \dfrac{-2}{3} \right)}^{2}}-7\left( \dfrac{-2}{3} \right)-6=0 \\\ & \Rightarrow \dfrac{4}{3}-\dfrac{4}{3}=0 \\\ \end{aligned}$$ Therefore, we can verify that the value for is $$x=3,-\dfrac{2}{3}$$.