Question

A natural number is greater than the other by 5. The sum of their squares is 73. Find the numbers.
NOTE-NEED TO BE SOLVE BY STEPS

Answer 1

Answer:

3 and 8

Step-by-step explanation:

Let the number be:x

Let the other number be:x+5

Sum of their squares:x²+(x+5²)=73

x²+x²+10x+25=73

2x²+10x+25=73

2x²+10x=73-25

2x²+10x-48=0

Divide the whole equation by 2

x²+5x-24=0

Factorise by quadratic formula:

a=1,b=5,c=-24

Put these values in the quadratic formula and you get x=3

So if x=3 then x+5=8

Answer 2

Answer:

3 and 8

Step-by-step explanation:

let the smaller number be x

the larger number will be x + 5

sum of squares of the numbers is 73

${x}^{2} + {(x + 5)}^{2} = 73 \\ {x }^{2} + {x}^{2} + 10x + 25 = 73 \\ 2 {x}^{2} + 10x - 48 = 0 \\dividing \: equation \: by \: 2 \\ {x}^{2} + 5x - 24 = 0 \\ \\ {x}^{2} + 8x - 3x - 24 = 0 \\ x(x + 8) - 3(x + 8) = 0 \\ (x + 8)(x - 3) = 0 \\ x + 8 = 0 \: \: \: \: \: \: \: \: \: x - 3 = 0 \\ x = - 8 \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 3 \\ we \: take \: the \: positive \: value \: of \: x$

the numbers are

x = 3

x + 5 = 3+5= 8

hope you get your answer