Chapter 4.5, Problem 62HP

To determine

To find: the meaning of completing the square using some description.

Answer to Problem 62HP

Constant $\text{a}$ is sometimes not $\text{1}$ but it can usually divide the whole equation by whatever $\text{a}$ is to make it $1$ before finding the $\text{k}$$\text{and}$ $\text{d}$ .

Explanation of Solution

Given:

  ${\text{ax}}^{\text{2}}\text{+bx+c}$ .

Concept used:

An expression obtained from the square of binomial equation is a perfect square trinomial is in the form of ${\text{ax}}^{\text{2}}\text{+bx+c}$ is said to be a perfect square, if only if it satisfies the condition ${\text{b}}^{\text{2}}\text{=4ac}$ .

The perfect squares trinomial formula is given as:

  ${\left(\text{ax}\right)}^{\text{2}}{\text{+2abx+b}}^{\text{2}}\text{=}{\left(\text{ax+b}\right)}^{\text{2}}$ .

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

Calculation:

Completing the square means getting a quadratic polynomial the form ${\text{ax}}^{\text{2}}\text{+bx+c}$ into something that looks like this:

  ${\left(\text{x+k}\right)}^{\text{2}}\text{+d}$

Where $\text{k}$ and $\text{d}$ are some constants. Multiplying this out:

  ${\left(\text{x+k}\right)}^{\text{2}}{\text{+d=x}}^{\text{2}}{\text{+2kx+k}}^{\text{2}}\text{+d}$ .

Comparing the equation with ${\text{ax}}^{\text{2}}\text{+bx+c}$ :

  $\text{2k=b}$

  $\text{k=}\frac{\text{b}}{2}$ .

  ${\text{k}}^{\text{2}}\text{+d=c}$ .

  ${\text{d=c-k}}^{\text{2}}$ .

$\text{a}$ is sometimes not $\text{1}$ but it can usually divide the whole equation by whatever $\text{a}$ is to make it $1$ before finding the $\text{k}$ $\text{and}$ $\text{d}$ . that’s all it needs to do, find the $\text{k}$ and $\text{d}$ to write in that form to complete the square.

Hence, $\text{a}$ is sometimes not $\text{1}$ but it can usually divide the whole equation by whatever $\text{a}$ is to make it $1$ before finding the $\text{k}$ $\text{and}$ $\text{d}$ .