Question

How do you solve $3{{x}^{2}}-2x=4$ using the quadratic formula?

Answer 1

In this question, we are given a quadratic equation to solve. We first have to write the given equation in the standard form of a quadratic equation, which is $a{{x}^{2}}+bx+c=0$. For this we need to take $4$ on the right hand side of the equation $3{{x}^{2}}-2x=4$ on the left hand side and our equation will be converted to the standard form. Then, as stated in the question, we have to use the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. From the standard form of the equation, we can note the values of the coefficients and substitute them into the quadratic formula to get the solutions of the given equation.

Complete step-by-step solution:
The quadratic equation given in the above question is
$3{{x}^{2}}-2x=4$
The standard form of a quadratic equation is written as $a{{x}^{2}}+bx+c=0$, that is, the right hand side of the quadratic equation must be zero. For this, we subtract $4$ from both sides of the above equation to get
$\begin{aligned} & \Rightarrow 3{{x}^{2}}-2x-4=4-4 \\\ & \Rightarrow 3{{x}^{2}}-2x-4=0.........(i) \\\ \end{aligned}$
The above question is directing us to use the quadratic formula for solving the given equation. We know that the quadratic formula is given as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}........(ii)$
From the equation (i) we can note the values of the coefficients as

& \Rightarrow a=3 \\\ & \Rightarrow b=-2 \\\ & \Rightarrow c=-4 \\\ \end{aligned}$$ Substituting these in the equation (ii) we get $\begin{aligned} & \Rightarrow x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 3 \right)\left( -4 \right)}}{2\left( 3 \right)} \\\ & \Rightarrow x=\dfrac{2\pm \sqrt{4+48}}{6} \\\ & \Rightarrow x=\dfrac{2\pm \sqrt{52}}{6} \\\ \end{aligned}$ Now, we know that $\sqrt{52}=2\sqrt{13}$. Substituting this in the above equation, we get $$\begin{aligned} & \Rightarrow x=\dfrac{2\pm 2\sqrt{13}}{6} \\\ & \Rightarrow x=\dfrac{1\pm \sqrt{13}}{3} \\\ & \Rightarrow x=\dfrac{1}{3}\pm \dfrac{\sqrt{13}}{3} \\\ \end{aligned}$$ **Hence, we have solved the given equation using the quadratic formula and the solutions are $x=\dfrac{1}{3}+\dfrac{\sqrt{13}}{3}$ and $x=\dfrac{1}{3}-\dfrac{\sqrt{13}}{3}$.** **Note:** The standard form of the quadratic equation is $a{{x}^{2}}+bx+c=0$ and not $a{{x}^{2}}+bx=c$. It must be remembered as the right hand side of the standard form of the quadratic equation is equal to zero. So we have to ensure the right hand side of the given quadratic equation is equal to zero. If it is not zero, then we have to make it zero like we have done in the above solution.