Question

The difference of two numbers is $2$, and the sum of their

squares is $10$. Find the numbers.

Short answer

The numbers are$3$and$1$or$-1$and$-3$.

Step-by-step solution

Step 1. Given information

Given that,

The difference of the two number is $2$. So, $a-b=2$.

The sum of their squares is $10$. So, $a^2+b^2=10$.

So, the system of equations are,

$\left\{\begin{array}{l}a-b=2\\ a^2+b^2=10\end{array}\right.$

Step 2 Let us write the first equation as,

$$\begin{gathered} a-b=2 \\ a=b+2 \end{gathered}$$

Substitute the value of $a$in the first equation.

$$\begin{gathered} {\left(b+2\right)}^2+b^2=10 \\ {b}^2+4b+4+b^2=10 \\ 2b^2+4b+4-10=0 \\ 2b^2+4b-6=0 \end{gathered}$$

Now factor out the common terms.

$$\begin{gathered} 2\left(b^2+2b-3\right)=0 \\ {b}^2+2b-3=0 \end{gathered}$$

Step 3 we can solve for b by using the quadratic formula -b±b2-4ac2a where a=1,b=2,c=-3.

$$\begin{gathered} b=\frac{-2\pm \sqrt{2^2-4\left(1\right)\left(-3\right)}}{2\left(1\right)} \\ b=\frac{-2\pm 4}{2} \\ b=\frac{-2+4}{2}orb=\frac{-2-4}{2} \\ b=\frac{2}{2}orb=\frac{-6}{2} \\ b=1orb=-3 \end{gathered}$$

Step 4 Substitute the value of b in the first equation.

Put $b=1$in the first equation.

$$\begin{gathered} a-1=2 \\ a=2+1 \\ a=3 \end{gathered}$$

Put $b=-3$in the first equation.

$$\begin{gathered} a-\left(-3\right)=2 \\ a+3=2 \\ a=2-3 \\ a=-1 \end{gathered}$$

So, the required numbers are,$3$ and$1$or$-1$and$-3$.