Question

use the appropriate method of solving the quadratic equation,y=2x^2-5x+1.

Answer 1

To solve the quadratic equation \( y = 2x^2 - 5x + 1 \), we can use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this equation, the coefficients are:
- \( a = 2 \)
- \( b = -5 \)
- \( c = 1 \)

Now, let's compute the discriminant (\( D \)):

\[
D = b^2 - 4ac = (-5)^2 - 4(2)(1) = 25 - 8 = 17
\]

Since the discriminant is positive, there will be two distinct real roots.

Now we can apply the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{17}}{2 \cdot 2}
\]

This simplifies to:

\[
x = \frac{5 \pm \sqrt{17}}{4}
\]

Thus, we have two possible solutions for \( x \):

1. \( x_1 = \frac{5 + \sqrt{17}}{4} \)
2. \( x_2 = \frac{5 - \sqrt{17}}{4} \)

These are the solutions for the equation \( y = 2x^2 - 5x + 1 = 0 \). If you need the approximate numerical values:

1. For \( x_1 \):

\[
x_1 \approx \frac{5 + 4.123}{4} \approx \frac{9.123}{4} \approx 2.28075
\]

2. For \( x_2 \):

\[
x_2 \approx \frac{5 - 4.123}{4} \approx \frac{0.877}{4} \approx 0.21925
\]

So the approximate solutions are \( x \approx 2.28 \) and \( x \approx 0.22 \).