Analyzing Errors When Solving Quadratic Equations

8.5.4 Analyzing Errors When Solving Quadratic Equations

8.5.4 • Analyzing Errors When Solving Quadratic Equations

Cool Down

Consider $(x - 5) (x + 1) = 7$ . Priya reasons that if this is true, then either $x - 5 = 7$ or $x + 1 = 7$ . So, the solutions to the original equation are 12 and 6.

Do you agree? If not, where was the mistake in Priya’s reasoning?

Compare your answer:

Disagree. Priya solved the equation using the reasoning we would use with the zero product property, but the zero product property only works if the product of two factors is 0. We can tell that 12 isn’t a solution because $(12 - 5) (12 + 1)$ is 91, not 7.

Consider $x^{2} - 10 x = 0$ . Diego says that to solve we can just divide each side by $x$ to get $x - 10 = 0$ , so the solution is 10. Mai says, “I wrote the expression on the left in factored form, which gives $x (x - 10) = 0$ , and ended up with two solutions: 0 and 10.”

Do you agree with either strategy? Be prepared to show your reasoning.

Compare your answer:

Agree with Mai’s strategy. By substituting 0 for $x$ and then 10 for $x$ , you can see that they are both solutions to the original equation. Diego’s strategy of dividing by a variable eliminates one of the solutions, leaving only one solution.

Consider the equation $x^{2} = - 16$ . Mona says that there are two solutions, $x = 4$ or $x = - 4$ .

Do you agree? If not, where was the mistake in Mona’s reasoning?

Compare your answer:

Disagree. There are no numbers we can square to get a negative product. The equation has no solutions.

Consider the equation $(x - 3) (x - 3) = 0$ . Dominic says that there are two solutions, $x = 3$ or $x = - 3$ .

Do you agree? If not, where was the mistake in Dominic’s reasoning?

Compare your answer:

Disagree. Only 3 makes the equation true. The solution is $x = 3$ .