Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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]


Sagot :

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]