Module 7 Quadratic Equations An earlier module dealt with first degree equations. In this module the study of equations continues with second degree equations in one variable. These are termed quadratic equations. A quadratic equation in x is one having the form ax 2+ bx+c=0, where a, b, and c are real constants and a 0. Example: x2 - 6x + 16 = 0 4x2 - 9 = 0 3x2 - x = 0 Solving Quadratic Equations by Method of Factoring If in the equation ax 2+ bx+c=0, the left side is factorable ( meaning, it can be expressed as a product), the value of x can be determined by equating each factor to zero and then solving the resulting linear equation. Solution by factoring uses the Zero Factor Property which states that if the product of two factors is 0 ( zero), then at least one of the factors is zero. That is, if ab = 0, then either a = 0 or b = 0. Examples: 1. 6x2 - 7x - 20 = 0 (2x - 5) (3x + 4) = 0 Factor the given. 2x - 5 = 0 Equate both factors to zero. 2x = 5 x = 5/2 2. 4x2 - 9x = 0 x(4x - 9) = 0 x = 0 3. 3x + 4 = 0 3x = -4 x = -4/3 Solving for the value of x. Do the same procedure as in example number 1. 4x - 9 = 0 4x = 9 x = 9/4 9x2 - 25 = 0 (3x + 5) (3x - 5) = 0 Do the same procedure as in 3x + 5 = 0 3x - 5 = 0 example number 1. x = -5/3 x = 5/3 An alternative method of solving the quadratic equation in example 3 is by using the Square Root Property which states that if x2 = b , then x= b x= b . 9x2 - 25 = 0 9x2 = 25 x2 = 25/9 25 x= 9 x= 5 3 Solving Quadratic Equations by the Method of Completing the Square Examine the following perfect square trinomials. 1. x2 + 8x + 16 = (x + 4)2 or 2. x2 - 10x + 25 = (x - 5)2 3. x2 + 5x + 25/4 = (x + 5/2)2 Observe that the last term in each perfect square trinomial is the square of onehalf the coefficient of the second term. 2 1. 8 16= 2 2. 8 2 2 16= () () 8 2 2 2 4. 10 25= 2 5. 25 5 = 4 2 16= 3. () ( ) 2 () Also, note that the first term is simply x2. Therefore, given the expression x2 + bx, to make it a perfect square, we add b 2 2 () . Examples: 1. x2 + 4x + 3 = 0 x2 + 4x = -3 x2 + 4x + 4 = -3 + 4 (x + 2)2 = 1 x + 2 = 1 Transpose 3 to the right. Add 4 to both sides. The expression at the left side is factored as the square of a binomial. Extract the square roots of sides, using plus and minus sign at the right side. x = -2 1 Transpose 2 to the right. x1 = -2 + 1 = -1 Two values of x are obtained. x2 = -2 - 1 = -3 2. x2 - 6x - 1 = 0 x2 - 6x = 1 2 x - 6x + 9 = 1+ 9 x2 - 6x + 9 = 10 (x - 3)2 = 0 x - 3 = 10 x = 3 3. 10 x1 = 3 + 10 x2 = 3 + 10 2x2 - 4x - 3 = 0 Dividing both members of the equation by 2. x2 - 2x - 3/2 = 0 x2 - 2x + 1 = 3/2 + 1 (x - 1)2 = 5/2 x-1= x=1 x=1 5 2 5 2 10 2 x 2 10 2 Solving Quadratic Equations by the Quadratic Formula The third way of solving quadratic equations is by the use of a formula obtained 2 ax + bx+ c=0 by solving the general quadratic equation a, b and c; a 0. The method of solving ax 2+ bx+ c=0 by completing the square is as follows: Divide both sides by a: b c x 2+ x+ =0 a a c Add a b c x 2+ x= a a Add to both sides: 2 [ ( )] b 1 /2 a for x in terms of the constants b b 2 c b2 x 2+ x+ 2 = + a a 4 a2 4a to both sides: Factor the left side: ( x+ b 2 c b2 = + 2a a 4 a2 ) Combine fractions at the right side into a single fraction: Extract the sqare root of both sides: ( x+ b 2 b24 ac = 2a 4 a2 ) ( b b24 ac x+ = 2a 4 a2 2 ) Transpose b 2a x= to the right: b b2 4 ac 2 2a 4a Simplify the right side: x= b b2 2ac 2a 2a Combine the fractions at the right side: x= b b 4 ac 2a 2 This result is known as the quadratic formula. b b24 ac x= 2a Examples: 1. 2x( x + 5) = 6 2x2 + 10x - 6 = 0 x + 5x - 3 = 0 Solution: a = 1 , b = 5 , c = -3 b b 4 ac 2a 2 x= 5 5 24 (1)(3) x= 2(1) x= 5 25+ 12 2 x 5 37 2 2. 3(3x2 + 4) = 2(2x) 9x2 - 4x + 12 = 0 Solution: a = 9, b = -4, c = 12 b b24 ac x= 2a x= (4) (4)24(9)(12) 2( 9) x= 4 16432 18 x= 4 416 18 x= 4 16(26) 18 x=26 4 4 26 18 is not a real number. 26 imaginary numbers where I is used to stand for belongs to a new set of numbers called 1 and i2 = -1. Thus and 26 = 1 44 i 26 18 26 . In example 2, the values of x are 4+ 4 i 26 18 . Discriminant and Roots In the quadratic formula, the quantity inside the radical sign, b 2 - 4ac determines the nature of the roots of the quadratic equation. The quantity b 2 - 4ac is called the discriminant of the quadratic equation 2 ax + bx+ c=0 . If a, b and c are integers, the nature of the roots is determined as follows: Discriminant 1. Positive and square of an integer 2. Positive but not the square of an integer 3. Zero 4. Negative Type of Roots Distinct and Rational Distinct and Irrational Equal and Rational Not real Examples: Determine the types of solutions for the following equations: 1. 6x2 + 5x - 4 = 0 a = 6, b = 5, c = -4 b2 - 4ac = 52 - 4(6)(-4) = 121 Since 121 = 112, then the solutions are two distinct rational numbers. 2. 2x2 - 4x + 1 = 0 a = 2, b = -4, c = 1 b2 - 4ac = (-4)2 - 4(2)(1) = 8 Since 8 is irrational ( 8 is not the square of an integer), then the solutions are two distinct irrational numbers. 3. 9x2 + 12x + 4 = 0 a = 9, b = 12, c = 4 b2 - 4ac = 122 - 4(9)(4) = 0 The discriminant is zero; hence the roots are equal and rational. 1. 2x2 + 3x + 5 = 0 a = 2, b = 3, c = 5 b2 - 4ac = 32 - 4(2)(5) = - 31 Since b2 - 4ac is negative then the roots are not real. Forming a Quadratic Equation Whose Roots are Given A quadratic equation whose roots are given maybe obtained by reversing the operation of solving quadratic equations by factoring. Example 1: Form a quadratic equation whose roots are 2 and -3/4. x = 2; x = -3/4 or 4x = -3 (x - 2) ( 4x + 3) = 0 4x2 - 5x - 6 = 0 Example 2: Form a quadratic equation whose roots are 2 x=2 + 3 ; x=2 3 . 3 . 3 ) ( x - 2 + 3 ) = 0 3 )] [( x - 2 + 3 )] = 0 (x - 2 [(x - 2 (x - 2)2 - ( 3 )2 = 0 (x - 2)2 - 3 = 0 x2 - 4x + 4 - 3 = 0 x2 - 4x + 1 = 0 4x2 - 5x - 6 = 0 Equations Containing Radicals Certain equations involving radicals lead to quadratic equations Example 1: 5 x +6=x+ 2 5x + 6 = ( x + 2 )2 square both sides 5x + 6 = x2 + 4x + 4 expand the right side x2 - x - 2 = 0 (x - 2 ) ( x + 1) = 0 x-2 =0 x=2 transpose all terms to the left factor the left side x + 1=0 x = -1 Check: For x = 2: 5(2)+ 6 For x = -1: = 2+2 5(1)+6 = -1 + 2 16 1=1 = 4 4 = 4 1=1 Example 2: 72 x - 3x 4x = 0 72 x - 3x= 4x ( leave only one radical in one side 72 x - 3x )2 = ( 4x )2 3x ( 72 x 2 72 x 3x+ ) = 4 - x 10 - 3x 2 ( 72 x ) (3x) = 4-x square both sides expand the left side combine like terms and multiply the two radicals at the left 6 - 2x (6 - 2x) 2 2 2113 x +2 x2 2113 x+ 2 x 2 4 )2 transpose the radical to the right square both sides 36 - 24x + 4x2 = 84 - 52x + 8x2 expand both the left and the right sides 4x2 - 28x + 48 = 0 transpose all terms to the left x2 - 7x + 12 = 0 (x-3)(x-4) = 0 Check: factor the left side x-3 = 0 x-4 = 0 x = 3 x = 4 For x = 3: For x = 4: 72(3) - 3(3) 4(3) 72(4) - 3(4 ) 4(4) = 0 11 1 - 0 - 1 = 0 0 = 0 = 0 -1 = 0 0 = 0 Example 3: x+ 4 - 2 x+1+1 = 0 x+ 4= 2 x +11 x+ 4 ( 2 =( 2 x +11 )2 x + 4 = 2x + 1 - 2 2 4( 2 x +1 2 x +1+1 = x -2 2 x +1 )2 = ( x - 2)2 8x + 4 = x2 - 4x + 4 x2 - 12x = 0 x( x - 12 ) = 0 x = 0 x - 12 = 0 x = 12 Check: For x = 0: 0+4 - 2(0)+1+1 For x = 12: = 0 12+ 4 - 2(12)+1+1 = 0 2 - 1 + 1 = 0 4 - 5 + 1 = 0 2 0 0 = 0 x = 12 satisfies the given radical equation, but x = 0 does not. Hence, 12 is the only root of the equation. We call 0 an extraneous root which is a root of the equation obtained by clearing off radicals, but not a root of the original equation. The procedure used for solving radical equations may be summarized as follows: 1. Arrange the terms of the equation so that a single radical appears on one side of the equation. 2. Eliminate the radical by squaring both sides of the equation. 3. Repeat steps 1 and 2, if necessary, until an equation that is free of any radical sign is obtained. 4. Solve the resulting equation. 5. Check each of the values obtained by substituting in the original equation. Reject any extraneous roots. Equations in Quadratic Form There are equations that are not quadratic but can be written in quadratic form. These are equations of the form au2 +bu+ c=0 , a 0. 2n n ax + bx + c=0 which can be transformed into These are equations in quadratic form. Example 1: x4 - 5x2 + 4 = 0 the given equation is quadratic in x2 ( x2 )2 - 5( x2 )1 + 4 = 0 Let u = x2 u2 - 5u + 4 = 0 substitute u for x2 and u2 for x4 (u-1)(u-4)=0 factor the left side u-1=0 u-4=0 u=1 u=4 x2 = 1 x2 = 4 x = 1 x = 2 Example 2: x2/3 + 7x1/3 + 12 = 0 the given equation is quadratic in x1/3 (x1/3 )2 + ( 7x1/3 )1 + 12 = 0 Let u = x1/3 u2 - 7u + 12 = 0 substitute u for x1/3 and u2 for x2/3 (u+3)(u+4)=0 factor the left side u+3=0 u+4=0 u = -3 u = -4 x1/3 = -3 x1/3 = -4 x = ( - 3 )3 x = ( -4 )3 x = - 27 x = - 64 Assignment 7 A. Solve the following quadratic equations by factoring. 1. 3x2 - 4x = 0 2. x2 - 100 = 0 3. x2 - 8x + 12 = 0 4. 8x2 + 3x = 9 - 3x 5. 15x2 + 5 = 13 + 14x 6. (3x + 2)(x + 1) = 10 7. (4x + 1)2 = 25 5 2 6 + = 8. x +3 x x+ 1 9. 10. 3x 2x 1 + = x +1 x3 x +1 2 x +4 3 4 x + = x+3 x x 2 +3 x B. Solve the following quadratic equations by the method of completing the square. 1. x2 - 4x - 12 = 0 2. x2 + 8x - 2 = 0 3. x2 + 3x + 1 = 0 4. y2 - 5y - 10 = 0 5. 2x2 + 2x - 11 = 0 6. (x - 4)2 = 100 7. (y + 2)2 = 25 8. (x + 3) (2x + 1) - 4 = 0 9. (2x + 1) (3x + 1) = 15 8 z 2 z 10. +1=0 3 C. Solve the following quadratic equations using quadratic formula. 1. x2 + 4x - 12 = 0 2. x2 + 6x + 5 = 0 3. 2x2 - 4x + 1 = 0 4. 2m ( m + 2) = 0 5. 4x2 = 3x - 1 6. 5x2 = 2 - 9x 7. (2x + 1) (3x + 1) = 0 8. (x + 3) (2x + 1) - 4 = 0 5 2 6 + = 9. x +3 x x+ 1 10. 1 1 1 = x1 x +1 x +1 D. Compute the discriminant to determine the type of roots of each of the following equations. 1. -30x = -9x2 - 15 2. 25x2 + 20x - 2 = 0 3. 18x2 + 13x - 12 = 0 4. 2x2 - x = 0 5. x2 - 4x + 4 = 0 E. Find a quadratic equation with integral coefficients whose roots are the indicated numbers. 1. 4 , -3 2. 6 , -2 3. , 4 4. 2/3 , -3 5. 1 + 21, 1+ 21 F. Find the roots. 2 1. x 7 = 3 2. 5 x +6 =x+2 3. 6 x8 =x 4. x+3 5. 2 x +13 - x+10 6. 54 x + 134 x 7. x+6 8. x 9. x2x 2 10. x23 x + 4 x+1 - = + =3 6 x+6 =1 - 3 x +4 = - x2 + x +7 - =4 x 6 x 12 x2x +3 =0 - 1 =0 G. Solve the following equations using quadratic form. 1. x4 - 10x2 + 9 = 0 2. 4x4 - 5x2 + 1 = 0 3. X6 - 9x3 + 8 = 0 4. 8x6 - 91x3 + 216 = 0 5. x - x1/2 - 30 = 0 6. x - 4x1/2 + 3 = 0 7. x-2 + x-1 - 12 = 0 8. x-4 - 8x-2 - 9 = 0 9. x4/3 - 5x2/3 + 4 = 0 10. x2/3 + x1/3 - 2 = 0 11. ( x2 - 5x )2 + ( x2 - 5x ) - 12 = 0 12. ( 2x2 + x )2 - 4( 2x2 + x ) + 3 = 0 2 x+1 x+ 1 2 7 15=0 13. x x ( ) ( ) 14. x1 2 x1 - 2=0 x x ( ) ( ) 15. x+ 1 x +1 x3 x3 1 /2 ( )( ) 2=0 -End