Chapter 4.5, Problem 41PPS

To determine

To find: the solution of the equation by completing the square.

Answer to Problem 41PPS

  $\text{x=2i+5}$ and $\text{x=-2i+5}$ .

Explanation of Solution

Given:

  ${\text{x}}^{\text{2}}\text{-10x+29=0}$ .

Concept used:

To solve the ${\text{ax}}^{\text{2}}\text{+bx+c=0}$ by completing the square:

1. Transform the equation so that the constant term $\text{c}$, is alone on the right side.
2. If $\text{a}$ ,the leading coefficient of ${\text{x}}^2$ term is not equal to $1$ , divide both side by $\text{a}$ .
3. Add the square of half the coefficient of the x term, ${\left(\frac{\text{b}}{\text{2a}}\right)}^{\text{2}}$ to both sides of the equation.
4. Factor the left side as the square of binomial.
5. Take the square root on both sides and remember: ${\left(\text{x+q}\right)}^{\text{2}}\text{=r}$ Is equivalent to $\left(\text{x+q}\right)\text{=±}\sqrt{\text{r}}$ .

Calculation:

  ${\text{x}}^{\text{2}}\text{-10x+29=0}$ .

  ${\text{x}}^{\text{2}}\text{-2}\left(\text{x}\right)\left(5\right)+{\left(5\right)}^2-{\left(5\right)}^2+29=0$ .

Using algebraic identities:

  ${\text{a}}^{\text{2}}{\text{-2ab+b}}^{\text{2}}\text{=}{\left(\text{a-b}\right)}^{\text{2}}$ .

  ${\left(\text{x}-5\right)}^2+4=0$ .

Taking square root on both side:

  $\left(\text{x-5}\right)\text{=}\pm \sqrt{\text{4}}\text{i}$ .

  $\left(\text{x-5}\right)\text{=}\pm \text{2i}$ .

Hence, $\text{x=2i+5}$ and $\text{x=-2i+5}$ .