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?
+122 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 \(\boxed{36}\).
+122 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 \(\boxed{36}\).
