Join the growing community of curious minds on IDNLearn.com. Join our community to receive timely and reliable responses to your questions from knowledgeable professionals.

Solve [tex]\(2x^2 - 8x = -7\)[/tex].

A. [tex]\(-2 \pm \sqrt{2}\)[/tex]
B. [tex]\(-2 \pm 2\sqrt{2}\)[/tex]
C. [tex]\(\frac{2 \pm \sqrt{2}}{2}\)[/tex]
D. [tex]\(2 \pm \frac{\sqrt{2}}{2}\)[/tex]


Sagot :

Certainly! Let's solve the quadratic equation [tex]\(2x^2 - 8x + 7 = 0\)[/tex] step by step.

1. Form the quadratic equation: The given equation can be expressed in standard form as
[tex]\[2x^2 - 8x + 7 = 0.\][/tex]

2. Identify the coefficients: For the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex],
- [tex]\(a = 2\)[/tex]
- [tex]\(b = -8\)[/tex]
- [tex]\(c = 7\)[/tex]

3. Use the quadratic formula: The quadratic formula states that the solutions to [tex]\(ax^2 + bx + c = 0\)[/tex] are given by:
[tex]\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\][/tex]

4. Plug in the coefficients: Substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the quadratic formula:
[tex]\[x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(7)}}{2(2)}.\][/tex]

5. Simplify the expression: Calculate the values inside the formula step by step.
- First, compute [tex]\(-b\)[/tex]:
[tex]\[-(-8) = 8.\][/tex]

- Next, compute [tex]\(b^2\)[/tex]:
[tex]\[(-8)^2 = 64.\][/tex]

- Then, compute [tex]\(4ac\)[/tex]:
[tex]\[4(2)(7) = 56.\][/tex]

- Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[64 - 56 = 8.\][/tex]

- Compute the square root of the discriminant:
[tex]\[\sqrt{8} = 2\sqrt{2}.\][/tex]

- Put all these into the quadratic formula:
[tex]\[x = \frac{8 \pm 2\sqrt{2}}{4}.\][/tex]

6. Simplify the final expression:
[tex]\[x = \frac{8}{4} \pm \frac{2\sqrt{2}}{4},\][/tex]
[tex]\[x = 2 \pm \frac{\sqrt{2}}{2}.\][/tex]

Therefore, the solutions to the equation [tex]\(2x^2 - 8x + 7 = 0\)[/tex] are:
[tex]\[x = 2 + \frac{\sqrt{2}}{2} \quad \text{and} \quad x = 2 - \frac{\sqrt{2}}{2}.\][/tex]

In decimal form, these solutions are approximately:
[tex]\[x \approx 2.7071 \quad \text{and} \quad x \approx 1.2929.\][/tex]