Question

How do you solve the quadratic equation by completing the square method $3{{x}^{2}}+9x-1=0$ ?

Answer 1

Hint: Now we are given with a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ . To solve the equation we will first divide the whole equation with a. Now we will add and subtract the term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ and then use ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to simplify the equation. Now we will take the square root on both sides and hence solve for x.

Complete step-by-step solution:
Now let us consider the given equation $3{{x}^{2}}+9x-1=0$ . We know that the given equation is a quadratic equation in one variable of the form $a{{x}^{2}}+bx+c=0$ . Now we want to find the solution of the given equation. Hence we want to find the value of x such that the equation holds for that value of x. Now to find the roots of the equation we will use the completing square method.
Now first comparing the given equation $3{{x}^{2}}+9x-1=0$ with the general quadratic equation $a{{x}^{2}}+bx+c$ we get a = 3, b = 9 and c = -1.
Now to use the completing square method we want the coefficient of ${{x}^{2}}$ to be 1. Hence we will divide the whole equation with a which is 3.
Hence we get the equation as ${{x}^{2}}+3x-1$ .
Now we will add and subtract the term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ in the equation to form a complete square.
Here we have b = 9 and a = 3 hence we will add and subtract ${{\left( \dfrac{9}{6} \right)}^{2}}$ Hence we get,
$\begin{aligned} & \Rightarrow 3{{x}^{2}}+9x-1+{{\left( \dfrac{9}{6} \right)}^{2}}-{{\left( \dfrac{9}{6} \right)}^{2}}=0 \\\ & \Rightarrow {{x}^{2}}+3x+\dfrac{81}{36}-\dfrac{81}{36}-1=0 \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow {{x}^{2}}+3x+\dfrac{81}{36}+\dfrac{-81-36}{36}=0 \\\ & \Rightarrow {{x}^{2}}+3x+\dfrac{81}{36}=\dfrac{117}{36} \\\ \end{aligned}$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Now using this we get
$\Rightarrow {{\left( x+\dfrac{3}{6} \right)}^{2}}=\dfrac{117}{36}$
Now taking square root on both sides we get,
$\begin{aligned} & \Rightarrow x+\dfrac{3}{6}=\dfrac{\pm \sqrt{117}}{6} \\\ & \Rightarrow x=\dfrac{-3\pm \sqrt{117}}{6} \\\ \end{aligned}$
Hence the roots of the given equation are $x=\dfrac{-3+\sqrt{117}}{6}$ and $x=\dfrac{-3-\sqrt{117}}{6}$

Note: Now note that for solving quadratic equations of the form $a{{x}^{2}}+bx+c=0$ we also have a direct formula. To find the roots we have $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . hence using this we can easily find the roots of the equation. Also note that the formula is derived from the completing square method itself.