Unit 4 Part 4.doc - Mat 108 Intermediate Algebra Unit 4...

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Mat 108 Intermediate AlgebraUnit 4 Part 4Quadratic Equations and the Quadratic FormulaApparently, a mathematician (maybe Bombelli in the 16thcentury) tired of solving quadratic equations by completing the squareand decided to shortcut this repetitive process. He followed the same process (completing the square) used for centuries but on the general quadratic 20axbxcwith coefficients of a, b, and crather than one specific example after another. The result of his work led to the following theorem: If a quadratic equation in one variable is in standard form(20axbxc) and the coefficients (a, b, and c) are any complex numbers, then the solutions are as follows:222444222bbacbbacbbacororaaaFurther more, if we call these solutions 1rand 2r, then 1212bcrrand rraa g. This second part of the theorem is often called the root testand can be used to check solutions without substituting them in the original problem.Let’s try Example 1:Solve 2323xxusing the quadratic formula. First, we must put the quadratic equation in standard form:2232323233230xxxxxxNext, we have to identify a, b, and c.For this example (now that it’s in standard form), we see that 3;2;3abc  and we substitute these into the formula: 2242224 332 3bbaca 1
Simplifying, we have 243624022 10402 10,666orsimplifying theaswe haveThe last expression is reducible (there’s a GCF of “2” in all three terms): 1103We believe that the solution set for this quadratic equation is 1103Using the second part of the theorem, we can check these answers without actually substituting them into the original equation. Let’s label our answers: 211011033rand r. If we can show that 212bcrrand rraa g(i.e. both are true), we will knowthat ouranswers are correct. Let’s try showing that 2brra (remember that 3;2;3abc  ): 121101102333rr(They have the same denominators and the two radicals would cancel.) Is this result (the23), the same as the result if we evaluate ba? Let’s see: 2233ba Yes it is. Now we have to multiply the two answers and verify that the result is the same as ca1211011011 101 10100110911339991rrFOILor gIsthis result the same as ca? Let’s see:
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