Question

Which is the solution set for the equation 2x^2-3x+1 =0

Answer 1

Given the equation

$2x^2-3x+1=0$

the solution can be found with the quadratic formula, as follows:

$\begin{gathered} x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot1}}{2\cdot2} \\ x_{1,2}=\frac{3\pm\sqrt[]{9^{}-8}}{4} \\ x_{1,2}=\frac{3\pm1}{4} \\ x_1=\frac{3+1}{4}=1 \\ x_2=\frac{3-1}{4}=\frac{1}{2} \end{gathered}$

The solution set is {1/2, 1}