Question

The square of the sum of two numbers is 144 and the sum of of their square is 180, find the number... Please I really need the answer now ​

Answer 1

Answer:

The numbers are either ±13.35 and ±1.35 or ±1.35 and ±13.3

Step-by-step explanation:

Let the two numbers be 'x' and 'y'.

Given:

The squares of the sum of two numbers = 144

The sum of their squares = 180

The sum of the numbers = $x+y$

The square of the sum of numbers = $(x+y)^2$

Square of first number = $x^2$

Square of second number = $y^2$

The sum of their squares = $x^2+y^2$

Now, as per question:

$(x+y)^2=144---(1)\\\\x^2+y^2=180---(2)$

Expanding equation (1) using the formula $(a+b)^2=a^2+b^2+2ab$

This gives,

$(x+y)^2=144\\\\x^2+y^2+2xy=144\\\\\textrm{Plug in the value of }x^2+y^2\textrm{ from equation (2)}\\\\180+2xy=144\\\\2xy=144-180\\\\2xy=-36\\\\xy=-\frac{36}{2}\\\\xy=-18\\\\y=-\frac{18}{x}---(3)$

Plug in the value of 'y' from equation (3) in equation (1). This gives,

$x^2+(-\frac{18}{x})^2=180\\\\x^2+\frac{324}{x^2}=180\\\\x^4+324=180x^2\\\\x^4-180x^2+324=0\\\\\textrm{On solving, we get:}\\\\x^2=178.2\ or\ x^2=1.8\\\\\textrm{Square root both sides, we get:}\\\\x=\pm13.35\ or\ x=\pm1.35$

Therefore,

$y=-\frac{18}{x}=-\frac{18}{\pm13.35} = \pm1.35\ or\\\\y=-\frac{18}{x}=-\frac{18}{\pm1.35}=\pm13.3$

Therefore, the numbers are either ±13.35 and ±1.35 or ±1.35 and ±13.3