Question

Find the real solutions, if any, of the equation$3x^2+x-1=0$.

Short answer

The solutions are$\left\{\frac{-1-\sqrt{13}}{6},\frac{-1+\sqrt{13}}{6}\right\}$.

Step-by-step solution

Step 1. Given Information

We are given the equation$3x^2+x-1=0$and we need to find the real roots of it.

Step 2. Finding the discriminant

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

Comparing it with the given equation we get,

$a=3,b=1,c=-1$.

Discriminant of the equation is

role="math" localid="1646502307564" $$\begin{gathered} b^2-4ac \\ ={\left(1\right)}^2-4\left(3\right)(-1) \\ =1+12 \\ =13>0 \end{gathered}$$

Thus there are two unequal real solutions for$x$.

Step 3. Finding the roots

Using the quadratic formula to find the roots:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values we get,

role="math" localid="1646502325187" $$\begin{gathered} x=\frac{-1\pm \sqrt{1^2-4\left(3\right)\left(-1\right)}}{2\left(3\right)} \\ \Rightarrow x=\frac{-1\pm \sqrt{1+12}}{6} \\ \Rightarrow x=\frac{-1\pm \sqrt{13}}{6} \\ \Rightarrow x=\frac{-1-\sqrt{13}}{6},\frac{-1+\sqrt{13}}{6} \end{gathered}$$