We have two numbers that we will call x and y.
One number exceeds the other by 5, so we can express it as an equation like:
[tex]y=x+5[/tex]where y exceeds x by 5.
The sum of the squares is 157, so we can express this as:
[tex]x^2+y^2=157[/tex]We can replace y with the first expression and obtain:
[tex]\begin{gathered} x^2+(x+5)^2=157 \\ x^2+x^2+10x+25=157 \\ 2x^2+10x+25-157=0 \\ 2x^2+10x-132=0 \\ x^2+5x-66=0 \end{gathered}[/tex]We now have a quadratic equation.
We can solve it as:
[tex]\begin{gathered} x=\frac{-5\pm\sqrt{5^2-4(1)(-66)}}{2(1)} \\ x=\frac{-5\pm\sqrt{25+264}}{2} \\ x=\frac{-5\pm\sqrt{289}}{2} \\ x=\frac{-5\pm17}{2} \\ \Rightarrow x_1=\frac{-5-17}{2}=\frac{-22}{2}=-11 \\ \Rightarrow x_2=\frac{-5+17}{2}=\frac{12}{2}=6 \end{gathered}[/tex]We have two possible solutions: -11 and 6.
We can check both:
1) If x = -11, then y is -6, and the sum of the squares is 121 + 36 = 157, so it is valid.
2) If x = 6, then y is 11 and the sum of the squares will also be 157, so it is also a valid solution.
Answer:
If both numbers are positive, they are 6 and 11.
If negative numbers are allowed, the pair of numbers -11 and -6 is a valid solution too.
