Question

Find the real solutions of the equation:

$6x^2-5x+1=0$

Short answer

The solution of the given equation is:

$$\begin{gathered} x=\frac{1}{2} \\ x=\frac{1}{3} \end{gathered}$$

Step-by-step solution

Step 1. Given information

The given equation is$6x^2-5x+1=0$

Step 2. Factorise the equation

Using the factorisation method, we get:

$$\begin{gathered} 6x^2-5x+1=0 \\ 6x^2-3x-2x+1=0 \\ 3x(2x-1)-(2x-1)=0[takingout3xand-1ascommon] \\ \left(2x-1\right)(3x-1)=0 \end{gathered}$$

Step 3. Evaluating the value

Using zero product property, we get:

$$\begin{gathered} \left(2x-1\right)=0 \\ 2x=1 \\ x=\frac{1}{2} \end{gathered}$$

or

$$\begin{gathered} 3x-1=0 \\ 3x=1 \\ x=\frac{1}{3} \end{gathered}$$

Step 4. Verify the solution

Substituting $x=\frac{1}{2}$in the given equation, we get:

$$\begin{gathered} 6x^2-5x+1=0 \\ 6{\left(\frac{1}{2}\right)}^2-5\left(\frac{1}{2}\right)+1=0 \\ 6\times \frac{1}{4}-5\times \frac{1}{2}+1=0 \\ \frac{3}{2}-\frac{5}{2}+1=0 \\ \frac{-2}{2}+1=0 \\ -1+1=0 \\ 0=0 \end{gathered}$$

This is true.

Substituting role="math" localid="1645888581245" $x=\frac{1}{3}$in the given equation, we get:

$$\begin{gathered} 6x^2-5x+1=0 \\ 6{\left(\frac{1}{3}\right)}^2-5\left(\frac{1}{3}\right)+1=0 \\ 6\times \frac{1}{9}-5\times \frac{1}{3}+1=0 \\ \frac{2}{3}-\frac{5}{3}+1=0 \\ \frac{2-5}{3}+1=0 \\ \frac{-3}{3}+1=0 \\ -1+1=0 \\ 0=0 \end{gathered}$$

This is true.

Thus the solution of the given function is:

role="math" localid="1645888988809" $$\begin{gathered} x=\frac{1}{2} \\ x=\frac{1}{3} \end{gathered}$$