Q. 74 - Precalculus Enhanced with Graphing Utilities

Question

The product of two numbers is $10$, and the difference of their squares is $21$. Find the numbers.

Step-by-Step Solution

Answer

The two numbers are $5 and 2$ or $-5 and -2$.

Step 1. Given information.

Assume the two numbers to be x and y.

From the question,

$$\begin{gathered} x\times y=10 ...... \left(i\right) \\ {x}^2-y^2=21 ...... \left(ii\right) \end{gathered}$$

Solve equation (i) for x,

$x=\frac{10}{y} ...... \left(iii\right)$

Substitute the value of x in equation (ii),

$$\begin{gathered} {\left(\frac{10}{y}\right)}^2-y^2=21 \\ \frac{100}{y^2}-y^2=21 \end{gathered}$$

Step 2. Find the two numbers.

Continuing the above equation,

$$\begin{gathered} 100-y^4-21y^2=0 \\ y=\pm 2 \end{gathered}$$

Substitute $y=2$ in equation (iii),

$$\begin{gathered} x=\frac{10}{2} \\ x=5 \end{gathered}$$

Substitute $y=-2$ in equation (iii),

$$\begin{gathered} x=\frac{10}{-2} \\ x=-5 \end{gathered}$$

Therefore, the numbers are $5 and 2$ or $-5 and -2$.