Week 6: Precalculus 11

Zero Product Law

This week in math we learned the zero product law.

The zero product law makes it easy and helps us when we are trying to solve quadratic equations.  The basics of zero product law is when, a -b = 0 then, a or b must equal 0.

At first I was confused about what a quadratic equation was and the difference between one and a linear equation.  Something that helps me understand the difference is that linear equations do not include exponents.

If my equation was \sqrt{2} – 81 = 0, since I know my perfect squares I notice that they both can be square rooted into the conjugates (x – 9) and (x + 9).  So my equation now looks like,                (x – 9)(x + 9) = 0.

From here using the zero product law (a -b = 0 then, a or b must equal 0.) I have to find two numbers to replace x in this case that will equal to 0 when added or subtracted by 9.

When I look at (x – 9) I know that x has to be 9 because 9 – 9 = 0.  For when I look at (x + 9) I know in the end it has to equal to 0 so, I know x has to equal -9 because -9 + 9 =0.

Finally, I have one x equal to 9 and the other equal to -9.  Now to verify I can replace x with either -9 or 9 to check if only one or both solutions work.  In this case, both solutions work because as long as one of your terms is equal to 0 it will be multiplied by the other and anything multiplied to 0 is 0.

 

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