Solving Quadratic Equations Using the Quadratic Formula

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1 Section 9 : Solving Quadratic Equations Using the Quadratic Fmula Quadratic Equations are equations that have an x term as the highest powered term. They are also called Second Degree Equations. The Standard Fm of a Second Degree Equation The polynomial Ax + Bx + C is set equal to 0 Ax + Bx + C = 0 The number in front of the x term is referred to with the letter A The number in front of the x term is referred to with the letter B The constant number is referred to with the letter C Note: The value of A should not be zero. If it is the equation will be a First Degree Equation not a Second Degree Equation. We learned how to solve First Degree Equations in past chapters. Solving Quadratic Equations In Chapter Six equations of the fm Ax + Bx + C = 0 were solved by the use of the Zero Fact Rule. This rule was used to solve equations that have products (facts) that are equal to zero. (x 3)(x + ) = 0 (x + )(x ) = 0 (x)(x 6) = 0 x (5x + ) = 0 The use of the Zero Fact Rule required that the second degree expression could be facted. If the second degree expression cannot be facted you do not want to solve by facting then another method can be used to solve the equation. This method can be used to solve ALL second degree equations. The Quadratic Fmula The Quadratic Fmula uses the values f A, B and C in the quadratic equation Ax + Bx + C = 0 and substitutes them into the Quadratic Fmula to find the two values f x that are solutions to the Second Degree Equation Ax + Bx + C = 0 If Ax + Bx + C = 0 then (B) ± ( B) (A)(C) (A) Note: The value of A should not be zero. If it is the equation will be a First Degree Equation not a Second Degree Equation. We learned how to solve First Degree Equations in past chapters. Math 100 Section 9 Page 1 01 Eitel

2 The Quadratic Fmula If Ax + Bx + C = 0 then (B) ± Note: The value of A should not be zero. ( B) (A)(C) (A) Steps to solve a second degree equation by the use of the Quadratic Fmula Step 1. Get the equation in the fm Ax + Bx + C = 0 Step. List the values f A, B and C Step 3. Find the value of the discriminate B AC Step. Put the values f B, A and B AC into the Quadratic Fmula. Step 5. Reduce the square root completely if possible and then reduce the remaining fraction if possible. Step 1. Get the equation in the fm Ax + Bx + C = 0 Get all the terms on one side of the = sign in the crect der Ax + Bx + C = 0. We will chose to do this in a way that keeps the Ax term positive. Many of the problems in this section will be set up in this fm at the start of the problem. If not, move the terms by addition subtraction. Example 1 Example Example 3 x + 9 x = 3x 5 7x 6 = 9x subtract 9 from both sides subtract 3x from both sides add 9x to both sides and add 5 to both sides x + x 9 = 9 9 x 3x + 5 = 3x 5 3x + 5 7x x + 9x x + x 9 = 0 x 3x + 5 = 0 7x 9x 6 = 0 Math 100 Section 9 Page 01 Eitel

3 Step. List the values f A, B and C in Ax + Bx + C = 0 Example 1 Example Example 3 x 7x + 9 = 0 A = B = 7 C = 9 3x + 6x 5 = 0 A = 3 B = 6 C = 5 x x + 3 = 0 A = B = 1 C = 3 Step 3. Find the value of the discriminate (B) (A)(C) The value under the radical sign is (B) (A)(C) and is called the discriminate. The value of (B) (A)(C) determines the type of number the solution will be and the amount of wk that will be required to solve the quadratic fmula f x. Example 1 Example Find (B) (A)(C) Find (B) (A)(C) f x x 10 = 0 x 5x + = 0 A = B = 1 C = 10 A = 1 B = 5 C = (B) (A)(C) ( 1) ()( 10) = = 1 (B) (A)(C) ( 5) (1)() = 5 = 17 Example 3 Example Find (B) (A)(C) Find (B) (A)(C) f x + 6x +1= 0 A = B = 6 C = 1 f x x + = 0 A = B = 5 C = (B) (A)(C) (6) ()(1) = = 0 (B) (A)(C) ( ) ()() = 3 = Math 100 Section 9 Page 3 01 Eitel

4 Step. Put the values f B, A and (B) (A)(C) into the Quadratic Fmula. Example 1 Example If 3x + 5x +1= 0 then A = 3 B = 5 C = 1 If x 3x 1= 0 then A = B = 3 C = 1 and (B) (A)(C) and (B) (A)(C) ( 5) (3)(1) 5 1 = 13 ( 3) ()( 1) 9 + = 17 (B) ± (B) (A)(C) (A) (B) ± (B) (A)(C) (A) (5) ± 13 (3) ( 3) ± 17 () 5 ± ± 17 The Quadratic Fmula If Ax + Bx + C = 0 then (B) ± (B) (A)(C) (A) Notice that there are two values f x that are solutions. The ± sign means that the term will be + in one version of the solution and in the other. Each of the separate equations gives a separate solution f x. It is common to start the solution process with a single equation with the ± sign and then separate the expression into the separate solutions at the end of the reducing process. (B) + (B) (A)(C) (A) (B) Example 1 Example (B) (A)(C) (A) 5 ± 17 can also be written as two separate solutions ± 11 can also be written as two separate solutions Math 100 Section 9 Page 01 Eitel

5 Putting It All Together The value under the radical sign is expressed as (B) (A)(C) and is called the discriminate. The value of (B) (A)(C) determines the type of number the solution will be and the amount of wk required to reduce the answer. The value f the discriminate (B) (A)(C) can be any one of four possible types of numbers: Case 1. When the discriminate (B) (A)(C) under the square root symbol is a positive number that cannot be reduced because it does not have a perfect square fact then the quadratic fmula cannot be reduced further. The solution will look like because the square root of 7 cannot be reduced Case. When the discriminate (B) (A)(C) under the square root is a negative number then the solution will not be a real number. STOP reducing the solution as soon as you see (B) (A)(C) is negative and write NRN. ± 15 = NRN 6 because the square root of 15 is not a real number Case 3. When the discriminate (B) (A)(C) under the square root is not a perfect square but the irrational number can be reduced to a radical with a smaller number under it because it has a perfect square fact then the radical must be reduced. That will leave you with an expression that may be reduced further. If all the three integers outside the radical sign have a common fact then reduce them by the common fact as the last step. The solution will look like Case. When the discriminate (B) (A)(C) under the square root is a perfect square then the solution will be two rational numbers. It will require several steps to find the two rational numbers. First replace the perfect square under the radical with its whole number value. Next break the equation with the ± sign into two separate fractions, one with a + sign and one with a sign. Then simplify each fraction. The solution will look like 1 3 Math 100 Section 9 Page 5 01 Eitel

6 Case 1: The discriminate is a positive number and the irrational number cannot be reduced at all. When the discriminate (B) (A)(C) under the square root symbol is a positive number that cannot be reduced because it does not have a perfect square fact then the quadratic fmula cannot be reduced further. (B) ± (B) (A)(C) (A) Example 1A Example 1B x 5x + = 0 x 3x = 0 A = 1 B = 5 C = A = B = 5 C = Find (B) (A)(C) ( 5) (1)() = 5 = 17 Find (B) (A)(C) ( 3) (3)( ) = 9 + = 33 ( 5) ± 17 3 ± 33 5 ± 17 = ( 3) ± 33 The 17 cannot be reduced at all. 17 does not have a perfect square fact a pair of facts. When this happens your answer is complete The 33 cannot be reduced at all. 33 does not have a perfect square fact a pair of facts. When this happens your answer is complete 5 ± 17 which can also be written 3 ± 33 which can also be written Math 100 Section 9 Page 6 01 Eitel

7 Case : The discriminate (B) (A)(C) under the square root is a negative number When the discriminate (B) (A)(C) under the square root is a negative number then the solution will not be a real number. STOP reducing the solution as soon as you see (B) (A)(C) is negative and write NRN. (B) ± (B) (A)(C) (A) Example A Example B 6x x +1= 0 5x + x + 3 = 0 A = 6 B = C = 1 A = 5 B = C = 3 Find (B) (A)(C) () (6)(1) = 16 = Find (B) (A)(C) () (5)(3) = 60 = 5 ( ) ± 1 () ± 10 5 If the final number under the radical(the discriminant) is a negitive number then NO REAL NUMBERS will wk and we write If the final number under the radical(the discriminant) is a negitive number then NO REAL NUMBERS will wk and we write NRN NRN Math 100 Section 9 Page 7 01 Eitel

8 Case 3. The discriminate is a positive number and the irrational number can be reduced to a smaller irrational number. When the discriminate (B) (A)(C) under the square root is not a perfect square but the irrational number can be reduced to a radical with a smaller number under it because it has a perfect square fact then the radical must be reduced. That will leave you with an expression that may be reduced further. If all the three integers outside the radical sign have a common fact then reduce them by the common fact as the last step. (B) ± (B) (A)(C) (A) Example 3A Example 3B x 6x +1= 0 A = B = 6 C = 1 Find B (A)(C) 36 ()(1) = = 0 x + x = 0 A =1 B = C = Find B (A)(C) 16 (1)( ) =16 + = ( 6) ± 0 reduce 0 = 5 ± reduce = 6 6 ± 5 all of the three integers outside the radical have a common fact ± 6 all of the three integers outside the radical have a common fact 6 / 3 ± / 1 5 / / ± / 1 6 / 1 3 ±1 5 ± Math 100 Section 9 Page 01 Eitel

9 Case : The discriminate is a Perfect Square When the discriminate (B) (A)(C) under the square root is a perfect square then the solution will be two rational numbers. It will require several steps to find the two rational numbers. First replace the perfect square under the radical with its whole number value. Next break the equation with the ± sign into two separate fractions, one with a + sign and one with a sign. Then simplify each fraction. (B) ± (B) (A)(C) (A) Example A Example B x + 5x + = 0 3x + 7x 6 = 0 A = 1 B = 5 C = A = 3 B = 7 C = 6 Find (B) (A)(C) (5) (1)() = 5 16 = 9 Find (B) (A)(C) (7) (3)( 6) = = 11 5 ± 9 reduce 9 (7) ± 11 (3) reduce 11 5 ± 3 7 ± = = 1 = = = 6 = = = Note: If the discriminate is a Perfect Square then the quadratic equation can be solved by facting. The homewk requires you to solve these type of quadratic equations by the use of the quadratic equation. On the test you can use either method. Math 100 Section 9 Page 9 01 Eitel

10 (B) ± (B) (A)(C) (A) Example C Example C Solved by the Solved by facting Quadratic Equation x x 10 = 0 A = B = 1 C = 10 x x 10 = 0 Fact the equation (x 5)(x + ) = 0 Find B (A)(C) (1) ( )( 10) = = 1 Set each fact equal to 0 x 5 = 0 x + = 0 ( 1) ± 1 () Solve both equations f x 5 1 ± 1 reduce 1 1 ± = 10 = = = 5 Note: If the discriminate is a Perfect Square then the quadratic equation can be solved by facting. The homewk requires you to solve these type of quadratic equations by the use of the quadratic equation. On the test you can use either method. Math 100 Section 9 Page Eitel

11 Special Cases In some quadratic equations the coefficient in front of the x term (the value f B) the constant term (the value f C ) will be 0. If B = 0 C = 0 then the Quadratic Equation is seldom used as a solution technique. It is much faster and easier to solve f x by facting. Examples 5A and 6A show show the facting method. The same problem is then solved by the use of the Quadratic Equation to show how the Quadratic Equation could be used to get the same solutions as the facting technique. Example 5A Example 6A What if B = 0 What if C = 0 x 9 = 0 Fact (x 3) (x + 3) = Example 5B x 9 = 0 A = B = 0 C = 9 6x 3 0 Fact 3x(x 1) = Example 6B 6x 3 0 A = 6 B = 3 C = 0 Find B (A)(C) 0 ()( 9) = 1 Find B (A)(C) 9 (6)(0) = 9 0 = 9 0 ± 1 0 ±1 reduce 1 ( 3) ± ± 3 1 reduce = 1 = = 6 1 = = 1 = = 0 1 = Math 100 Section 9 Page Eitel

Example 1 Example 2 Example 3. Example 4 Example 5

Example 1 Example 2 Example 3. Example 4 Example 5 Section 7 5: Solving Radical Equations Radical Equations are equations that have a radical expression in one or more of the terms in the equation. Most of the radicals equations in the section will involve