Question

solve for x in the equation $2x^2 - 5x + 1 = 3$.
\\( \circ \\) $x = \frac{5}{2} \pm \frac{\sqrt{29}}{2}$
\\( \circ \\) $x = \frac{5}{2} \pm \frac{\sqrt{41}}{4}$
\\( \circ \\) $x = \frac{5}{4} \pm \frac{\sqrt{29}}{2}$
\\( \circ \\) $x = \frac{5}{4} \pm \frac{\sqrt{41}}{4}$

Explanation:

Step1: Rearrange the equation

Subtract 3 from both sides to get the quadratic equation in standard form \(ax^2 + bx + c = 0\).
\(2x^2 - 5x + 1 - 3 = 0\)
Simplify to \(2x^2 - 5x - 2 = 0\). Here, \(a = 2\), \(b = -5\), \(c = -2\).

Step2: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\).
Substitute \(a = 2\), \(b = -5\), \(c = -2\) into the formula:
First, calculate the discriminant \(b^2 - 4ac = (-5)^2 - 4\times2\times(-2)=25 + 16 = 41\).
Then, \(x=\frac{-(-5)\pm\sqrt{41}}{2\times2}=\frac{5\pm\sqrt{41}}{4}=\frac{5}{4}\pm\frac{\sqrt{41}}{4}\).

Answer:

\(x=\frac{5}{4}\pm\frac{\sqrt{41}}{4}\) (the fourth option, assuming the options are ordered from top to bottom as first, second, third, fourth)