Use the quadratic formula, [tex]$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$[/tex], to solve the equation [tex]$2x^2 - 8x + 7 = 0$[/tex]. Round to the nearest hundredths place.

A. [tex]x=-2.71[/tex] and [tex]x=-1.29[/tex]

B. [tex]x=1.29[/tex] and [tex]x=2.71[/tex]

C. [tex]x=-5.25[/tex] and [tex]x=9.25[/tex]

D. [tex]x=5.17[/tex] and [tex]x=10.83[/tex]



Answer :

To solve the quadratic equation [tex]\(2x^2 - 8x + 7 = 0\)[/tex] using the quadratic formula, we need to use the given formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].

First, let's identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] from the equation:
[tex]\[a = 2\][/tex]
[tex]\[b = -8\][/tex]
[tex]\[c = 7\][/tex]

Next, we calculate the discriminant, which is given by [tex]\(b^2 - 4ac\)[/tex]:

[tex]\[ \text{Discriminant} = (-8)^2 - 4 \cdot 2 \cdot 7 \][/tex]
[tex]\[ = 64 - 56 \][/tex]
[tex]\[ = 8 \][/tex]

Now, we apply the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{\text{Discriminant}}}{2a} \][/tex]

Substituting in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and the discriminant:

[tex]\[ x = \frac{-(-8) \pm \sqrt{8}}{2 \cdot 2} \][/tex]

Simplifying the numerator and denominator:

[tex]\[ x = \frac{8 \pm \sqrt{8}}{4} \][/tex]

The square root of 8 can be simplified further:

[tex]\[ \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2} \][/tex]

Substituting this back into the equation:

[tex]\[ x = \frac{8 \pm 2\sqrt{2}}{4} \][/tex]

This can be simplified by dividing each part of the numerator by 4:

[tex]\[ x = \frac{8}{4} \pm \frac{2\sqrt{2}}{4} \][/tex]
[tex]\[ x = 2 \pm \frac{\sqrt{2}}{2} \][/tex]

We need the approximate numerical values rounded to the nearest hundredths.

Calculating the two possible values:

[tex]\[ x_1 = 2 + \frac{\sqrt{2}}{2} \approx 2.71 \][/tex]
[tex]\[ x_2 = 2 - \frac{\sqrt{2}}{2} \approx 1.29 \][/tex]

Thus, the solutions to the equation [tex]\(2x^2 - 8x + 7 = 0\)[/tex] are:

[tex]\[ x = 2.71 \quad \text{and} \quad x = 1.29 \][/tex]

Therefore, the correct answer is:

[tex]\[ x = 1.29 \quad \text{and} \quad x = 2.71 \][/tex]