Question

Solve by completing the square X^3 - 8x + 4 = -3

Answer 1

To solve the equation \( x^3 - 8x + 4 = -3 \), we first simplify it:

\[
x^3 - 8x + 4 + 3 = 0
\]

This simplifies to:

\[
x^3 - 8x + 7 = 0
\]

Next, we can use the method of completing the square for the expression \( x^3 - 8x \). However, it is worth noting that this equation is cubic and can also be approached using synthetic division or numerical methods.

To solve \( x^3 - 8x + 7 = 0 \):

1. **Identify if there is a rational root** using the Rational Root Theorem. The possible rational roots can be the factors of \( 7 \) (constant term) over the factors of \( 1 \) (leading coefficient), which gives us the candidates: \( \pm 1, \pm 7 \).

2. **Test these candidates**:
- For \( x = 1 \):
\[
1^3 - 8(1) + 7 = 1 - 8 + 7 = 0
\]
\( x = 1 \) is a root.

3. **Perform polynomial long division** to factor out \( (x - 1) \):
\[
x^3 - 8x + 7 = (x - 1)(x^2 + x - 7)
\]

4. **Now we need to solve for** \( x^2 + x - 7 = 0 \). We can apply the quadratic formula here:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1, b = 1, c = -7 \):
\[
x = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-7)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 28}}{2} = \frac{-1 \pm \sqrt{29}}{2}
\]

5. **Therefore, the solutions to the original equation \( x^3 - 8x + 7 = 0 \) are**:
\[
x = 1, \quad x = \frac{-1 + \sqrt{29}}{2}, \quad x = \frac{-1 - \sqrt{29}}{2}
\]

Summarizing, the solutions are:
\[
x = 1, \quad x = \frac{-1 + \sqrt{29}}{2}, \quad x = \frac{-1 - \sqrt{29}}{2}
\]