Q. 73 - Precalculus Enhanced with Graphing Utilities

Question

The product of two numbers is $4$, and the sum of their squares is $8$. Find the numbers.

Step-by-Step Solution

Answer

The two numbers are $2 and 2$ or $2 and 2$.

Step 1. Given information.

Assume the two numbers to be x and y.

From the question,

$$\begin{gathered} x\times y=4 ...... \left(i\right) \\ {x}^2+y^2=8 ...... \left(ii\right) \end{gathered}$$

Solve equation (i) for x,

$x=\frac{4}{y} ...... \left(iii\right)$

Step 2. Substitute values in equation (ii).

Substitute the value of x in equation (ii),

$$\begin{gathered} {\left(\frac{4}{y}\right)}^2+y^2=8 \\ \frac{16}{y^2}+y^2=8 \\ \frac{16}{y^2}+y^2-8=0 \\ 16+y^4-8y^2=0 \\ {\left(y^2-4\right)}^2=0 \end{gathered}$$

Step 3. Find the two numbers.

Take square root both the sides,

$$\begin{gathered} \sqrt{{\left(y^2-4\right)}^2}=0 \\ {y}^2-4=0 \\ y=\pm 2 \end{gathered}$$

Substitute $y=2$ in equation (iii),

$$\begin{gathered} x=\frac{4}{2} \\ x=2 \end{gathered}$$

Substitute $y=-2$ in equation (iii),

$$\begin{gathered} x=\frac{4}{-2} \\ x=-2 \end{gathered}$$

Therefore, the numbers are $2 and 2$ or $-2 and -2$.