Question

Solve the quadratic equation $x^2+1=\frac{13x}{6}$ by using the quadratic formula.

Answer 1

Determining the value of $x$ by using the quadratic formula

The standard form of the quadratic equation is $a{x}^2+bx+c=0$.

$$\begin{gathered} x^2+1=\frac{13x}{6} \\ \Rightarrow x^2-\frac{13x}{6}+1=0 \end{gathered}$$

The quadratic formula is $\begin{array}{rcl}x& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$.

For the given equation $x^2-\frac{13x}{6}+1=0$, the value of variables are $a=1,b=-\frac{13}{6}$ and $c=1$.

Substituting the values of $a,b$ and $c$ in quadratic formula,

$$\begin{gathered} \Rightarrow x=\frac{-(-{\displaystyle \frac{13}{6}})\pm \sqrt{{(-{\displaystyle \frac{13}{6}})}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)} \\ \Rightarrow x=\frac{{\displaystyle \frac{13}{6}}\pm \sqrt{{\displaystyle \frac{169}{36}}-4}}{2} \\ \Rightarrow x=\frac{{\displaystyle \frac{13}{6}\pm \sqrt{\frac{25}{36}}}}{2} \\ \Rightarrow x=\frac{{\displaystyle \frac{13}{6}}\pm {\displaystyle \frac{5}{6}}}{2} \\ \Rightarrow x=\frac{13\pm 5}{12} \\ \Rightarrow x=\frac{13-5}{12},\frac{13+5}{12} \end{gathered}$$

$$\begin{gathered} \Rightarrow x=\frac{8}{12},\frac{18}{12} \\ \therefore x=\frac{2}{3},\frac{3}{2} \end{gathered}$$

Therefore, the value of $x$ is $\frac{2}{3}$ or $\frac{3}{2}$.