Section 10 1: Solving Quadratic Equations Using the Quadratic Fmula Quadratic Equations are equations that have an x term as the highest power. They are also called Second Degree Equations. The Standard Fm of a Second Degree Equation 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 cannot be zero. If it is then the equation will not be a Quadratic Equation. 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 + 4) = 0 (x + )(x ) = 0 (4x)(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 expression whether they can be facted not. The Quadratic Fmula uses the values f A, B and C in the equation Ax + Bx + C = 0 and substitutes them into the fmula B± B 4AC A to find the two values f x that are solutions to the Second Degree Equation Ax + Bx + C = 0 The Quadratic Fmula If Ax + Bx + C = 0 then B± B 4AC A Note: The value of A cannot be zero. Math 100 Section 10 1 Page 1 010 Eitel
The Quadratic Fmula If Ax + Bx + C = 0 then B± B 4AC A Note: The value of A cannot be zero. 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 4AC Step 4. Put the values f B, B 4AC and A into the Quadratic Fmula. Step 5. Reduce the square root completely 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. 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 to one side by addition subtraction. Example 1 Example Example 3 5x + 4 9 subtract 9 from both sides x + 4x 9 = 9 9 5x = 3x 5 subtract 3x from both sides and add 5 to both sides x 3x + 5 = 3x 5 3x + 5 7x + 6 = 9x add 9x to both sides 7x + 6 + 9 9x + 9x x + 4x 9 = 0 x 3x + 5 = 0 x + 9x + 6 = 0 Math 100 Section 10 1 Page 010 Eitel
Step. List the values f A, B and C in Ax + Bx + C = 0 Example 1 Example Example 3 x 7x + 9 = 0 3x + 6x 5 = 0 4x x + 3 = 0 A = B = 7 C = 9 A = 3 B = 6 C = 5 A = 4 B = 1 C = 3 Step 3. Find the value of the discriminate B 4AC The value under the radical sign is B 4(A)(C) and is called the discriminant. The value of B 4(A)(C) determines the type of number the solution will be and the amount of wk required to solve the quadratic fmula f x. Example 1 Example Find B 4(A)(C) Find B 4(A)(C) f x x 10 = 0 A = B = 1 C = 10 1 4()( 10) =1+ 0 = 1 x 5x + = 0 A =1 B = 5 C = Find B 4(A)(C) 5 4(1)() = 5 =17 Example 3 Example 4 Find B 4(A)(C) Find B 4(A)(C) f 4x + 6x +1 = 0 f 4x 5x + = 0 A = 4 B = 6 C =1 A = 4 B = 5 C = Find B 4(A)(C) 36 4(4)(1) = 36 16 = 0 Find B 4(A)(C) 5 4(4)() = 5 3 = 7 Math 100 Section 10 1 Page 3 010 Eitel
Step 4. Put the values f B, B 4AC and A into the Quadratic Fmula. Example 1 Example If 4x 6x +1= 0 then A = 4 B = 6 and B 4(A)(C) = 0 If x + 4 x 1 = 0 then A =1 B = 4 and B 4(A)(C) = 64 B± B 4AC A ( 6) ± 0 (4) 6 ± 0 B± B 4AC A (4)± 64 (1) 4 ± 64 The Quadratic Fmula If Ax + Bx + C = 0 then B± B 4AC A B + B 4AC A B B 4AC 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 expressions gives a separate solution f x. It is common to start the solution process with a single expression with the ± sign and then separate the expression into the separate solutions as part of the reducing process. Example 1 Example 5 ± 17 which can also be written as two separate solutions 4 ± 11 which can also be written as two separate solutions 5 + 17 5 17 4 + 11 4 11 Math 100 Section 10 1 Page 4 010 Eitel
Putting It All Together The value under the radical sign is expressed as B 4(A)(C) and is called the discriminant. The value of B 4(A)(C) determines the type of number the solution will be and the amount of wk required to solve the quadratic fmula f x. The value f the discriminant B 4(A)(C) can be any one of four possible types of numbers: Case 1. When the discriminant B 4(A)(C) 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 4 Case. When the discriminant B 4(A)(C) is a square root 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 5 + 7 3 5 7 3 Case 3. When the discriminant B 4(A)(C) 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 also be reduced. If all the three integers outside the radical have a common fact then reduce them by the common fact as the last step. The solution will look like 1+ 5 7 1 5 7 Case 4. When the discriminant B 4(A)(C) is a negative number under the square root then the solution will not be a real number. Stop reducing the solution as soon as you see b 4(a)(c) is negative and write NRN. Math 100 Section 10 1 Page 5 010 Eitel
Case 1: The discriminant is a Perfect Square 4 ± 15 = NRN 6 When the discriminant B 4(A)(C) 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. Example 1 Example x + 5x + 4 = 0 x + 4x 1 = 0 A =1 B = 5 C = 4 A =1 B = 4 C = 1 Find B 4(A)(C) 5 4(1)(4) = 5 16 = 9 Find B 4(A)(C) 16 4(1)( 1) =16 + 4 = 64 5± 9 reduce 9 4 ± 64 reduce 64 5± 3 4 ± 5+ 3 5 3 1 4 = = 1 = = 4 4 + 4 6 = 4 = = 1 = 6 Math 100 Section 10 1 Page 6 010 Eitel
In many quadratic equations the coefficient in front of the x term (the value f A) will be a number other than 1. In the examples below the value f A is greater than 1. B± B 4AC A Example 3 Example 4 x x 10 = 0 3x + 7x 6 = 0 A = B = 1 C = 10 A = 3 B = 7 C = 6 Find B 4(A)(C) 1 4( )( 10) =1+ 0 = 1 Find B 4(A)(C) 49 4(3)( 6) = 49 + 7 =11 ( 1) ± 1 () (7)± 11 (3) reduce 11 1± 1 4 reduce 1 7± 11 6 reduce 11 1± 9 4 7±11 6 1+ 9 = 10 4 = 5 7+11 6 = 4 6 = 3 1 9 4 = 4 = 7 11 = 1 6 6 = 3 5 3 3 Math 100 Section 10 1 Page 7 010 Eitel
In many 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 4x 9 = 0 Fact (x 3) (x + 3) = 0 6x 3 0 Fact 3x(x 1) = 0 3 3 0 1 Example 5B Example 6B Math 100 Section 10 1 Page 010 Eitel
4x 9 = 0 6x 3 0 A = 4 B = 0 C = 9 A = 6 B = 3 C = 0 Find B 4(A)(C) 0 4(4)( 9) =144 Find B 4(A)(C) 9 4(6)(0) = 9 0 = 9 0 ± 144 reduce 144 3 ± 9 1 reduce 9 0 ±1 3 ± 3 1 0 +1 = 1 = 3 3 + 3 1 = 6 1 = 1 0 1 = 1 = 3 3 3 1 = 0 1 = 0 3 3 1 0 Case : The discriminant cannot be reduced at all When the discriminant B 4(A)(C) is a square root that cannot be reduced at all because it does not have a perfect square fact then the quadratic fmula cannot be reduced further. Example 7 Example Math 100 Section 10 1 Page 9 010 Eitel
x 5x + = 0 4x 3x = 0 A =1 B = 5 C = A = 4 B = 5 C = Find B 4(A)(C) 5 4(1)() = 5 =17 Find B 4(A)(C) 9 4(3)( ) = 9 + 4 = 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 5 + 17 5 17 3 + 33 3 33 Case 3. The discriminant is not a perfect square but the irrational number can be reduced When the discriminant B 4(A)(C) 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 also be reduced. If all the three integers outside the radical have a common fact then reduce them by the common fact as the last step. Math 100 Section 10 1 Page 10 010 Eitel
Example 9 Example 10 4x 6x +1 = 0 A = 4 B = 6 C =1 Find B 4(A)(C) 36 4(4)(1) = 36 16 = 0 x + 4x = 0 A =1 B = 4 C = Find B 4(A)(C) 16 4(1)( ) =16 + = 4 6 ± 0 reduce 0 = 4 5 4 ± 4 reduce 4 = 4 6 6 ± 5 4 ± 6 all of the three integers outside the radical have a common fact 6 / 3 ± / 1 5 / 4 all of the three integers outside the radical have a common fact / 4 ± / 1 6 / 1 3 ±1 5 4 ±1 6 1 3 + 5 4 3 5 4 + 6 6 Case 4: The discriminant B 4(A)(C) is a negative number When the discriminant B 4(A)(C) is a negative number under the square root then the solution will not be a real number. Stop reducing the solution as soon as you see b 4(a)(c) is negative and write NRN. Math 100 Section 10 1 Page 11 010 Eitel
Example 11 Example 1 6x 4x +1 = 0 5x + x + 3 = 0 A = 6 B = 4 C =1 A = 5 B = C = 3 Find B 4(A)(C) 16 4(6)(1) =16 4 = Find B 4(A)(C) 4 4(5)(3) = 4 60 = 54 If the final number under the radical(the discriminant) reduces to a negitive number then NO REAL NUMBERS will wk and we write If the final number under the radical(the discriminant) reduces to a negitive number then NO REAL NUMBERS will wk and we write NRN NRN Math 100 Section 10 1 Page 1 010 Eitel