Solving Quadratic Equations with the Zero Product Property
Algebra 1
Practice
Problem 1
If the equation \((x+10) x=0\) is true, which statement is also true according to the zero product property?
Only \(x = 0\).
Either \(x = 0\) or \(x + 10 = 0\).
Either \(x^2 = 0\) or \(10x=0\).
Only \(x + 10 = 0\).
Problem 2
What are the solutions to the equation \((10-x)(3x-9)=0\)?
-10 and 3
-10 and 9
10 and 3
10 and 9
Problem 3
Solve each equation.
\((x-6)(x+5)=0\)
\((x-3)(\frac23 x - 6)=0\)
\((\text-3x-15)(x+7)=0\)
Problem 4
Consider the expressions \((x-4)(3x-6)\) and \(3x^2 - 18x + 24\).
Show that the two expressions define the same function.
Problem 5
Kiran saw that if the equation \((x+2)(x-4)=0\) is true, then by the zero product property, either \(x+2\) is 0 or \(x-4\) is 0. He then reasoned that, if \((x+2)(x-4)=72\) is true, then either \(x+2\) is equal to 72 or \(x-4\) is equal to 72.
Andre wants to solve the equation \(5x^2-4x-18=20\). He uses a graphing calculator to graph \(y=5x^2-4x-18\) and \(y=20\) and finds that the graphs cross at the points \((\text-2.39, 20)\) and \((3.19, 20)\).
Substitute each \(x\)-value that Andre found into the expression \(5x^2-4x-18\). Then evaluate the expression.
Why did neither solution make \(5x^2-4x-18\) equal exactly 20?