Q3. - Algebra 1

Question

Use the Quadratic formula to find the exact solutions of the equation $\mathbf{2}{x}^2-\mathbf{6}x+\mathbf{3}=\mathbf{0}$.

$$\begin{gathered} A\frac{\mathbf{3}\pm \sqrt{3}}{2} \\ B\frac{\mathbf{3}\pm \sqrt{2}}{4} \\ C\frac{\mathbf{2}\pm \sqrt{5}}{3} \\ D\frac{\mathbf{5}\pm \sqrt{2}}{2} \end{gathered}$$

Step-by-Step Solution

Answer

The exact solutions $2x^2-6x+3=0$ are $\frac{\mathbf{3}\pm \sqrt{3}}{2}$.

Step 1. Identify.

Identify the coefficients of the equations by comparing the given equation $a{x}^2+bx+c=0$.

 

$a=2,b=-6$ and $c=3$

Step 2. Write a quadratic formula.

A quadratic formula is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Step 3. Substitute and simplify.

Substitute 2 for a, $-6$ for b and 3 for c in $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$$\begin{gathered} x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\times 2\times 3}}{2\left(2\right)} \\ =\frac{12\pm \sqrt{36-24}}{4} \\ =\frac{3\pm \sqrt{3}}{2} \end{gathered}$$ 

 

Therefore, the exact solutions $2x^2-6x+3=0$ are $\frac{3+\sqrt{3}}{2}$ and $\frac{3-\sqrt{3}}{2}$. The obtained answer completely matches with option A, therefore option A is the required answer.