Question

One number exceeds another by 5 the sum of their square is 157. what are the number

Answer 1

Start by calling the first number as a variable:

`x`

then, since this number exceeds the other number by 5, the other number is:

`x-5`

then, write the equation that represents the sum of their square,

`\begin{gathered} x^2+(x-5)^2=157 \\ x^2+x^2-10x+25=157 \\ 2x^2-10x+25=157 \\ 2x^2-10x-132=0 \end{gathered}`

solve using the quadratic:

`\begin{gathered} x=(-(-10)\pm√((-10)^2-4(2)(-132)))/(2(2)) \\ x=(10\pm34)/(4) \\ x_1=(10+34)/(4) \\ x_1=11 \\ x_2=(10-34)/(4) \\ x_2=-6 \end{gathered}`

Answer:

since the result gives two answers there are two possible combinations that fulfill this statement;

in the first case 11 and 6, for the second case -11 and -6.