Certainly! Let's solve the quadratic equation [tex]\( x^2 + 4x - 7 = 0 \)[/tex] by completing the square and give the answers correct to two decimal places.
1. Start with the original equation:
[tex]\[
x^2 + 4x - 7 = 0
\][/tex]
2. Move the constant term to the other side of the equation to isolate the quadratic and linear terms:
[tex]\[
x^2 + 4x = 7
\][/tex]
3. To complete the square, take half of the coefficient of [tex]\( x \)[/tex], square it, and add it to both sides of the equation. The coefficient of [tex]\( x \)[/tex] is 4, so half of it is 2 and squaring it gives [tex]\( 2^2 = 4 \)[/tex]:
[tex]\[
x^2 + 4x + 4 = 7 + 4
\][/tex]
4. Simplify the equation:
[tex]\[
(x + 2)^2 = 11
\][/tex]
5. Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[
x + 2 = \pm \sqrt{11}
\][/tex]
6. Isolate [tex]\( x \)[/tex] by subtracting 2 from both sides:
[tex]\[
x = -2 \pm \sqrt{11}
\][/tex]
7. Calculate the values for [tex]\( x \)[/tex]. We have two solutions:
- For [tex]\( x = -2 + \sqrt{11} \)[/tex]:
[tex]\[
x \approx -2 + 3.31662 = 1.31662
\][/tex]
- For [tex]\( x = -2 - \sqrt{11} \)[/tex]:
[tex]\[
x \approx -2 - 3.31662 = -5.31662
\][/tex]
8. Therefore, the solutions to the equation [tex]\( x^2 + 4x - 7 = 0 \)[/tex] are:
[tex]\[
x \approx 1.32 \quad \text{and} \quad x \approx -5.32
\][/tex]
So, the solutions are [tex]\( x = 1.31662 \)[/tex] and [tex]\( x = -5.31662 \)[/tex], correct to two decimal places.
