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QUADRATIC EQUATIONS MSJC ~ San Jacinto Campus Math Center Workshop Series Janice Levasseur Basics • A quadratic equation is an equation equivalent to an equation of the type ax2 + bx + c = 0, where a is nonzero • We can solve a quadratic equation by factoring and using The Principle of Zero Products If ab = 0, then either a = 0, b = 0, or both a and b = 0. Ex: Solve (4t + 1)(3t – 5) = 0 Notice the equation as given is of the form ab = 0 set each factor equal to 0 and solve 4t + 1 = 0 Subtract 1 Divide by 4 4t = – 1 t=–¼ 3t – 5 = 0 3t = 5 t = 5/3 Add 5 Divide by 3 Solution: t = - ¼ and 5/3 t = {- ¼, 5/3} Ex: Solve x2 + 7x + 6 = 0 Quadratic equation factor the left hand side (LHS) x2 + 7x + 6 = (x + 6 )(x + 1 ) x2 + 7x + 6 = (x + 6)(x + 1) = 0 Now the equation as given is of the form ab = 0 set each factor equal to 0 and solve x+6=0 x=–6 x+1=0 x=–1 Solution: x = - 6 and – 1 x = {-6, -1} Ex: Solve x2 + 10x = – 25 Quadratic equation but not of the form ax2 + bx + c = 0 Add 25 x2 + 10x + 25 = 0 Quadratic equation factor the left hand side (LHS) x2 + 10x + 25 = (x + 5 )(x + 5 ) x2 + 10x + 25 = (x + 5)(x + 5) = 0 Now the equation as given is of the form ab = 0 set each factor equal to 0 and solve x+5=0 x+5=0 x=–5 x=–5 Solution: x = - 5 x = {- 5} repeated root Ex: Solve 12y2 – 5y = 2 Quadratic equation but not of the form ax2 + bx + c = 0 Subtract 2 12y2 – 5y – 2 = 0 Quadratic equation factor the left hand side (LHS) ac method a = 12 and c = – 2 ac = (12)(-2) = - 24 factors of – 24 that sum to - 5 1&-24, 2&-12, 3&-8, . . . 12y2 – 5y – 2 = 12y2 + 3y – 8y – 2 = 3y(4y + 1) – 2(4y + 1) = (3y – 2)(4y + 1) 12y2 – 5y – 2 = 0 12y2 – 5y – 2 = (3y - 2)(4y + 1) = 0 Now the equation as given is of the form ab = 0 set each factor equal to 0 and solve 3y – 2 = 0 4y + 1 = 0 3y = 2 4y = – 1 y = 2/3 y=–¼ Solution: y = 2/3 and – ¼ y = {2/3, - ¼ } Ex: Solve 5x2 = 6x Quadratic equation but not of the form ax2 + bx + c = 0 Subtract 6x 5x2 – 6x = 0 Quadratic equation factor the left hand side (LHS) 5x2 – 6x = x( 5x – 6 ) 5x2 – 6x = x(5x – 6) = 0 Now the equation as given is of the form ab = 0 set each factor equal to 0 and solve 5x – 6 = 0 x=0 5x = 6 x = 6/5 Solution: x = 0 and 6/5 x = {0, 6/5} Solving by taking square roots • An alternate method of solving a quadratic equation is using the Principle of Taking the Square Root of Each Side of an Equation If x2 = a, then x = + a Ex: Solve by taking square roots 3x2 – 36 = 0 First, isolate x2: 3x2 – 36 = 0 3x2 = 36 x2 = 12 Now take the square root of both sides: x 12 x 12 2 x 223 x2 3 Ex: Solve by taking square roots 4(z – 3)2 = 100 First, isolate the squared factor: 4(z – 3)2 = 100 (z – 3)2 = 25 Now take the square root of both sides: (z 3) 25 2 z 3 25 z–3=+5 z=3+5 z = 3 + 5 = 8 and z = 3 – 5 = – 2 Ex: Solve by taking square roots 5(x + 5)2 – 75 = 0 First, isolate the squared factor: 5(x + 5)2 = 75 (x + 5)2 = 15 Now take the square root of both sides: ( x 5 )2 15 x 5 15 x 5 15 x 5 15 , x 5 15 Completing the Square • Recall from factoring that a Perfect-Square Trinomial is the square of a binomial: Perfect square Trinomial x2 + 8x + 16 x2 – 6x + 9 Binomial Square (x + 4)2 (x – 3)2 • The square of half of the coefficient of x equals the constant term: ( ½ * 8 )2 = 16 [½ (-6)]2 = 9 Completing the Square • Write the equation in the form x2 + bx = c • Add to each side of the equation [½(b)]2 • Factor the perfect-square trinomial x2 + bx + [½(b)] 2 = c + [½(b)]2 • Take the square root of both sides of the equation • Solve for x Ex: Solve w2 + 6w + 4 = 0 by completing the square First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1. w2 + 66w = – 4 Add [½(b)]2 to both sides: b = 6 [½(6)]2 = 32 = 9 w2 + 6w + 9 = – 4 + 9 w2 + 6w + 9 = 5 (w + 3)2 = 5 Now take the square root of both sides ( w 3) 5 2 w 3 5 w 3 5 w {3 5,3 5} Ex: Solve 2r2 = 3 – 5r by completing the square First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1. 2r2 + 5r = 3 r2 + (5/2)r (5/2) = (3/2) Add [½(b)]2 to both sides: b = 5/2 [½(5/2)]2 = (5/4)2 = 25/16 r2 + (5/2)r + 25/16 = (3/2) + 25/16 r2 + (5/2)r + 25/16 = 24/16 + 25/16 (r + 5/4)2 = 49/16 Now take the square root of both sides (r 5 / 4) 49 /16 2 r 5 / 4 (7 / 4) r (5 / 4) (7 / 4) r = - (5/4) + (7/4) = 2/4 = ½ and r = - (5/4) - (7/4) = -12/4 = - 3 r = { ½ , - 3} Ex: Solve 3p – 5 = (p – 1)(p – 2) Is this a quadratic equation? FOIL the RHS 3p – 5 = p2 – 2p – p + 2 3p – 5 = p2 – 3p + 2 p2 – 6p + 7 = 0 Collect all terms A-ha . . . Quadratic Equation complete the square p2 – 6p = – 7 p2 – 6p + 9 = – 7 + 9 (p – 3)2 = 2 [½(-6)]2 = (-3)2 = 9 (p 3) 2 2 p3 2 p 3 2 p {3 2,3 2} The Quadratic Formula • Consider a quadratic equation of the form ax2 + bx + c = 0 for a nonzero • Completing the square ax bx c b c x x a a b b c b x x a 4a a 4a 2 2 2 2 2 2 2 The Quadratic Formula b b 4ac b x x a 4a 4a 4a b b 4ac x 2a 4a 2 2 2 2 2 2 2 2 2 Solutions to ax2 + bx + c = 0 for a nonzero are 2 b b 4ac x 2a Ex: Use the Quadratic Formula to solve1x2 + 7x 7 +6=0 Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by b b 2 4ac x 2a Identify a, b, and c in ax2 + bx + c = 0: a=1 b= 7 c= 6 Now evaluate the quadratic formula at the identified values of a, b, and c 7 7 2 4(1)(6) x 2(1) 7 49 24 x 2 7 25 x 2 75 x 2 x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6 x = { - 1, - 6 } Ex: Use the Quadratic Formula to solve 2 2m2 + 1m – 10 = 0 Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by b b 2 4ac m 2a Identify a, b, and c in am2 + bm + c = 0: a=2 b= 1 c = - 10 Now evaluate the quadratic formula at the identified values of a, b, and c 1 12 4(2)(10) m 2(2) 1 1 80 m 4 1 81 m 4 1 9 m 4 m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2 m = { 2, - 5/2 } Any questions . . .