Question

the difference of the square of two numbers is 45 the square of the smallest number is 4 times the larger number determines the small number

Answer 1

6

Let x is large number, y is smaller number

y^2 = 4x

According to problem,

x^2 - y^2 = 45

or, x^2 - 4x = 45

or, x^2 -4x -45 = 0

or, (x-9)(x+5) = 0

Either, x - 9 = 0

=> x = 9                     OR,

                                               X + 5 = 0

                                            => X = -5

x /= -5 as value of y cannot be negative. So, x = 9

therefore, y^2 = 4x

or, y^2 = 36 or, y = 6

Answer 2

$let \: the \: larger \: number \\ \: be \: x \: and \: smaller \: be \: y \\ {x}^{2} - {y}^{2} = 45 \\ also \: {y}^{2} = 4x \\ therefore \\ {x}^{2} - 4 x = 45 \\ {x}^{2} + 5x - 9x - 45 = 0 \\x (x + 5) - 9(x + 5) = 0 \\ (x + 5)(x - 9) \\ x = 9 \: and \: - 5 \\ x \: cant \: be \: - 5 \: becoz \\ \: then \: {y}^{2} \: will \: become \\ \: negative \\ y = 6$
hope it is your solution