Question

find a number whose square diminished by 119 is equal to ten times the excess of the number over 8

Answer 1

$\textbf{Answer}$

Suppose the required number is X

So the square of the number = X^2

Square of number diminished by 119 = X^2 - 119

Excess of X over 8 = X - 8

$\textbf{So according to the question,}$

X^2 - 119 = 10 (X - 8)

=> X^2 - 119 = 10X - 80

=> X^2 - 10X - 119 + 80 = 0

=> X^2 - 10X - 39 = 0

=> X^2 - 13X + 3X - 39 = 0

=> X(X - 13) + 3(X - 13) = 0

=> (X - 13) (X + 3) = 0

=> $\textbf{X = 13 or X = -3}$


$\textbf{So the required numbers are 13 and -3}$

$\textbf{Hope It Helps}$

$\textbf{Thanks}$