Solve for \(x\) in the equation \(2x^2 - 5x + 1 = 3\).

A. \(x = \frac{5}{2} \pm \frac{\sqrt{29}}{2}\)

B. \(x = \frac{5}{2} \pm \frac{\sqrt{41}}{4}\)

C. \(x = \frac{5}{4} \pm \frac{\sqrt{29}}{2}\)

D. [tex]\(x = \frac{5}{4} \pm \frac{\sqrt{41}}{4}\)[/tex]



Answer :

To solve the quadratic equation \( 2x^2 - 5x + 1 = 3 \), follow these steps:

1. Rewrite the equation in standard form:
[tex]\[ 2x^2 - 5x + 1 = 3 \implies 2x^2 - 5x + 1 - 3 = 0 \implies 2x^2 - 5x - 2 = 0 \][/tex]

2. Identify the coefficients \( a \), \( b \), and \( c \):
The quadratic equation is now in the standard form \( ax^2 + bx + c = 0 \) with:
[tex]\[ a = 2, \quad b = -5, \quad c = -2 \][/tex]

3. Calculate the discriminant \( \Delta \):
[tex]\[ \Delta = b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot (-2) = 25 + 16 = 41 \][/tex]

4. Apply the quadratic formula:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]

Substitute \( a \), \( b \), and \( \Delta \):
[tex]\[ x = \frac{-(-5) \pm \sqrt{41}}{2 \cdot 2} = \frac{5 \pm \sqrt{41}}{4} \][/tex]

5. Write the solutions:
[tex]\[ x = \frac{5}{4} \pm \frac{\sqrt{41}}{4} \][/tex]

So the correct answer is:
[tex]\[ x = \frac{5}{4} \pm \frac{\sqrt{41}}{4} \][/tex]