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Algebra 1

8.5.4 Analyzing Errors When Solving Quadratic Equations

Algebra 18.5.4 Analyzing Errors When Solving Quadratic Equations

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1.

Consider (x5)(x+1)=7(x5)(x+1)=7. Priya reasons that if this is true, then either x5=7x5=7 or x+1=7x+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?

2.

Consider x210x=0x210x=0. Diego says that to solve we can just divide each side by xx to get x10=0x10=0, so the solution is 10. Mai says, “I wrote the expression on the left in factored form, which gives x(x10)=0x(x10)=0, and ended up with two solutions: 0 and 10.”

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

3.

Consider the equation x2=16x2=16. Mona says that there are two solutions, x=4x=4 or x=4x=4.

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

4.

Consider the equation (x3)(x3)=0(x3)(x3)=0. Dominic says that there are two solutions, x=3x=3 or x=3x=3.

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

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