The difference between two numbers is 9, and the sum of the squares of each number is 158. What is the value of the product of the two numbers?

Guest Dec 29, 2020

#1**+1 **

__Solution #1:__ If the numbers are called \(x\) and \(y\), then \(y = x - 9\), and so the sum of squares

\(x^2+ y^2 = x^2 + (x-9)^2 = 153\)

\(x^2 + (x^2 - 18x + 81) = 153\)

\(2x^2 - 18x - 72 = 0\)

\(x^2 - 9x - 36 = 0\)

This is factorable. However, if you use the quadratic formula rather than just factoring, the formula shows that either \(x=12\) and \(y=3\), or \(x=-3\) and \(y=-12\). Either way, \(x\cdot y = \boxed{36}\).

*Solution #2:* \(x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\rightarrow2xy = 153–81 =72\text{,} \text{ ie., } xy = \boxed{36}.\)

Both ways, you get the same answer of \(\boxed{36}\).

cryptoaops Dec 29, 2020

#1**+1 **

Best Answer

__Solution #1:__ If the numbers are called \(x\) and \(y\), then \(y = x - 9\), and so the sum of squares

\(x^2+ y^2 = x^2 + (x-9)^2 = 153\)

\(x^2 + (x^2 - 18x + 81) = 153\)

\(2x^2 - 18x - 72 = 0\)

\(x^2 - 9x - 36 = 0\)

This is factorable. However, if you use the quadratic formula rather than just factoring, the formula shows that either \(x=12\) and \(y=3\), or \(x=-3\) and \(y=-12\). Either way, \(x\cdot y = \boxed{36}\).

*Solution #2:* \(x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\rightarrow2xy = 153–81 =72\text{,} \text{ ie., } xy = \boxed{36}.\)

Both ways, you get the same answer of \(\boxed{36}\).

cryptoaops Dec 29, 2020