This post about solving quadratic equations is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.
Find the values of x that satisfy this equation:
It’s a quadratic. That means the first step to solving it is to get everything on the left-hand side, leaving 0 on the right. In this case, that means that you subtract 5 from each side:
Now see if you can solve this quadratic by factoring: Look for two numbers that multiply to -15 and add up to 2. 5 and -3 fit the bill. So one factor is and one is . Then
And we said
Two things multiplied together can equal 0 only when at least one of those things equals 0. So to find solutions to (x + 5)(x – 3) = 0, set x + 5 equal to 0 and set x – 3 equal to 0. Then the factor theorem says that a polynomial that has a factor has a root at c, and a polynomial that has a root at c has a factor :
So two values of x satisfy the equation: x=-5, 3. To check this answer, substitute these values for x into the equation. If you get a true equation, your answer is correct.
The Accuplacer sample question to which this question is similar is multiple choice. If you are unable to solve by factoring, you can plug in answer choices until you find one that works.
Note: Not every quadratic can be solved by factoring. When you can’t factor, you can complete the square or use the quadratic formula. This post shows you how to complete the square. The steps from completing the square to solving the equation are left to you.
For more about solving a quadratic, see “How to Factor a Quadratic” and “How Can You Find a Quadratic’s Equation from Its Graph?”
The Accuplacer sample problem on which this one is based is #9 in the sample questions for the Accuplacer Advanced Algebra and Functions test.