Question

How do you solve $\dfrac{4}{3}{{x}^{2}}-2x+\dfrac{3}{4}=0$ using the quadratic formula?

Answer 1

Hint: We first have to simplify the given equation so as to get free from the fractions which is making the equation cumbersome. For this we multiply the equation by $12$ to obtain the simplified quadratic equation $16{{x}^{2}}-24x+9=0$. The quadratic formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. On substituting the values of the coefficients from the equation $16{{x}^{2}}-24x+9=0$, which are $a=16$, $b=-24$, and $c=9$, we will obtain the required solutions of the given equation.

Complete step-by-step answer:
The equation to be solved is given in the question as
$\dfrac{4}{3}{{x}^{2}}-2x+\dfrac{3}{4}=0$
Since it contains fractions, we have to simplify it. For this, we have to multiply the above equation with the LCM of the denominator numbers $3$ and $4$, which is equal to $12$. Therefore, multiplying both sides of the above equation by $12$ we get
\[\begin{align}
  & \Rightarrow 12\left( \dfrac{4}{3}{{x}^{2}}-2x+\dfrac{3}{4} \right)=12\left( 0 \right) \\
 & \Rightarrow \dfrac{48}{3}{{x}^{2}}-24x+\dfrac{36}{4}=0 \\
 & \Rightarrow 16{{x}^{2}}-24x+9=0......(i) \\
\end{align}\]
As stated in the question, we have to solve the given equation using the quadratic formula which is given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the equation (i) we note the values of the coefficients as $a=16$, $b=-24$, and $c=9$. Substituting these in the above formula, we get
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -24 \right)\pm \sqrt{{{\left( -24 \right)}^{2}}-4\left( 16 \right)\left( 9 \right)}}{2\left( 16 \right)} \\
 & \Rightarrow x=\dfrac{24\pm \sqrt{576-576}}{32} \\
 & \Rightarrow x=\dfrac{24\pm 0}{32} \\
 & \Rightarrow x=\dfrac{24}{32},\dfrac{24}{32} \\
 & \Rightarrow x=\dfrac{3}{4},\dfrac{3}{4} \\
\end{align}$
Hence, the solution of the given quadratic equation is $x=\dfrac{3}{4}$ and $x=\dfrac{3}{4}$.

Note: The simplification of the quadratic equation is not necessary. We can solve it with the coefficients as fractions as well. But for easing out the calculations, we simplified the given equation. Also, although we got a single solution to the given quadratic equation, we have to write the solution two times. This is because a quadratic equation has two solutions and therefore even if its roots are repeated, they are written two times.