Question

name:
to earn full credit, answer the following to the best of your ability in the first five minutes of class:
can you find two numbers that when added together they equal 7 and when multiplied together they equal 12?
can you find two numbers that when added together they equal 4 and when multiplied together they equal 3?

Explanation:

Step1: Set up equations for first - pair

Let the two numbers be \(x\) and \(y\). We have \(x + y=7\), so \(y = 7 - x\), and \(xy = 12\). Substitute \(y = 7 - x\) into \(xy = 12\), getting \(x(7 - x)=12\).
\[x(7 - x)=12\Rightarrow7x - x^{2}=12\Rightarrow x^{2}-7x + 12 = 0\]

Step2: Solve the quadratic equation

Factor the quadratic equation \(x^{2}-7x + 12 = 0\). We know that \(x^{2}-7x + 12=(x - 3)(x - 4)=0\). So \(x=3\) or \(x = 4\). When \(x = 3\), \(y=7 - 3=4\); when \(x = 4\), \(y=7 - 4 = 3\).

Step3: Set up equations for second - pair

Let the two numbers be \(m\) and \(n\). We have \(m + n=4\), so \(n = 4 - m\), and \(mn = 3\). Substitute \(n = 4 - m\) into \(mn = 3\), getting \(m(4 - m)=3\).
\[m(4 - m)=3\Rightarrow4m - m^{2}=3\Rightarrow m^{2}-4m + 3 = 0\]

Step4: Solve the second quadratic equation

Factor the quadratic equation \(m^{2}-4m + 3 = 0\). We know that \(m^{2}-4m + 3=(m - 1)(m - 3)=0\). So \(m = 1\) or \(m = 3\). When \(m = 1\), \(n=4 - 1=3\); when \(m = 3\), \(n=4 - 3 = 1\).

Answer:

For the sum of 7 and product of 12, the numbers are 3 and 4. For the sum of 4 and product of 3, the numbers are 1 and 3.