# IA In-Class 4-4 - Mat 108 Intermediate Algebra Unit 4 Part...

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Mat 108 Intermediate Algebra Unit 4 Part 4 Quadratic Equations and the Quadratic Formula Apparently, a mathematician (maybe Bombelli in the 16 th century) tired of solving quadratic equations by completing the square and decided to shortcut this repetitive process. He followed the same process ( completing the square ) used for centuries but on the general quadratic 2 0 ax bx c + + = with coefficients of a , b , and c rather 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 ( 2 0 ax bx c + + = ) and the coefficients ( a , b , and c ) are any complex numbers , then the solutions are as follows: 2 2 2 4 4 4 2 2 2 b b ac b b ac b b ac or or a a a - + - - - - - - Further more, if we call these solutions 1 r and 2 r , then 1 2 1 2 b c r r and r r a a + =- = g . This second part of the theorem is often called the root test and can be used to check solutions without substituting them in the original problem. Let’s try Example 1: Solve 2 3 2 3 x x = + using the quadratic formula . First, we must put the quadratic equation in standard form : 2 2 3 2 3 2 3 2 3 3 2 3 0 x x x x x x = + - - - - - - = Next, we have to identify a , b , and c . For this example (now that it’s in standard form ), we see that 3; 2; 3 a b c = =- =- and we substitute these into the formula: ( ) ( ) ( ) ( ) ( ) 2 2 4 2 2 2 4 3 3 2 3 b b ac a - - - - - - - 1
Simplifying, we have ( ) 2 4 36 2 40 2 2 10 40 2 10, 6 6 6 or simplifying the as we have + The last expression is reducible (there’s a GCF of “2” in all three terms): 1 10 3 . We believe that the solution set for this quadratic equation is 1 10 3 Using the second part of the theorem, we can check these answers without actually substituting them into the original equation. Let’s label our answers: 1 2 1 10 1 10 3 3 r and r + - = = . If we can show that 1 2 1 2 b c r r and r r a a + =- = g (i.e. both are true), we will know that our answers are correct. Let’s try showing that 1 2 b r r a + =- (remember that 3; 2; 3 a b c = =- =- ): 1 2 1 10 1 10 2 3 3 3 r r � � + - + = + = � � � � � � (They have the same denominators and the two radicals would cancel.) Is this result (the 2 3 ), the same as the result if we evaluate b a - ? Let’s see: ( ) ( ) 2 2 3 3 b a - - - Yes it is.
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