Question

an equation has solutions of m = -5 and m = 9. which could be the equation?
○ (m + 5)(m - 9) = 0
○ (m - 5)(m + 9) = 0
○ m² - 5m + 9 = 0
○ m² + 5m - 9 = 0

Explanation:

Step1: Recall Zero - Product Property

If a product of two factors is zero, then at least one of the factors is zero. That is, if \((x - a)(x - b)=0\), then \(x = a\) or \(x = b\).

Step2: Analyze the roots \(m=-5\) and \(m = 9\)

For the root \(m=-5\), we can rewrite it as \(m+5 = 0\) (by adding 5 to both sides of \(m=-5\)). For the root \(m = 9\), we can rewrite it as \(m - 9=0\) (by subtracting 9 from both sides of \(m = 9\)).

Step3: Form the equation using Zero - Product Property

Using the zero - product property, if \(m + 5=0\) and \(m-9 = 0\) are the factor equations, then the quadratic equation is \((m + 5)(m - 9)=0\).
Let's check the other options:

• For \((m - 5)(m+9)=0\), the roots would be \(m = 5\) and \(m=-9\) (since \(m - 5=0\Rightarrow m = 5\) and \(m + 9=0\Rightarrow m=-9\)), which does not match the given roots.
• For \(m^{2}-5m + 9=0\), we can use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b=-5\), \(c = 9\). The discriminant \(\Delta=b^{2}-4ac=(-5)^{2}-4\times1\times9=25 - 36=- 11<0\), so there are no real roots, let alone \(m=-5\) and \(m = 9\).
• For \(m^{2}+5m-9=0\), using the quadratic formula with \(a = 1\), \(b = 5\), \(c=-9\), the discriminant \(\Delta=5^{2}-4\times1\times(-9)=25 + 36 = 61\), and the roots are \(m=\frac{-5\pm\sqrt{61}}{2}\), which are not equal to \(-5\) and \(9\).

Answer:

\((m + 5)(m - 9)=0\) (the first option)