QUADRATIC EQUATIONS MSJC San Jacinto Campus Math Center Workshop Series Janice Levasseur
2
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.
3
Ex: Solve 4t+l (3t-5) = O Notice the equation as given is of the form -Y set each factor equal to 0 and solve 3t-5=O t = 5/3 Solution: Subtract 1 Divide by 4 Add 5 Divide by 3 1/4 and t = 1/4, 5/3}
4
Ex: solve + 7 x + 6 = O Quadratic equation -5 factor the left hand side (LHS) Now the equation as given is of the form -5 set each factor equal to 0 and solve ab = O Solution: x 8+1-0 = 6 and —1 * x
5
Ex: Solve + 10x = - 25 Quadratic equation but not of the form ax2 + bx + c = 0 Add 25 + 1 Ox + 25 = O Quadratic equation -5 factor the left hand side (LHS) + + 25 = (x + 5 5 ) + I Ox + 25 = (x + + 5) = O Now the equation as given is of the form ab = O -5 set each factor equal to 0 and solve 8+6e@ Solution: x 8+5-0 = _ 5 ...+ x = 5} repeated root
6
Ex: Solve - = 2 Quadratic equation but not of the form ax2 + bx + c = 0 Subtract 2 1 2y2 — -z? = 0 Quadratic equation -5 factor the left hand side (LHS) ac method -Y a = and c 12 - - 24 -Y factors of — 24 that sum to - 5 1&-24, 2&-12, 3&-8, o -5y-2= 12y2 + - 8y-2 = 3y(4y + l) — 2(4y + l) = (3y — 2)(4y + 1)
7
...> 1 2y2 — — 2 = O 12y2 - = (3y - + l) = O Now the equation as given is of the form ab = O -Y set each factor equal to 0 and solve 3y = 2 Y = 2/3 Solution: y = 2/3 and — {2/3, - }
8
Ex: Solve = Quadratic equation but not of the form ax2 + bx + c = 0 Subtract 6x — = 0 Quadratic equation -5 factor the left hand side (LHS) 5x2 — 6x = x( 5x Now the equation as given is of the form ab = O -5 set each factor equal to 0 and solve Solution: x 5x-6=o = O and 6/5 x
9
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 = a, then x = ± a
10
Ex: Solve by taking square roots — 36 = O First, isolate — 36 = 0 = 36 = 12 Now take the square root of both sides: x 12 12 2,203
11
- IOO Ex: Solve by taking square roots 4(z First, isolate the squared factor: = IOO (z — = 25 Now take the square root of both sides: 5=-2 Z 25 25 z = 3+5 8 and z = 3—
12
Ex: Solve by taking square roots 5(x + First, isolate the squared factor: + = 75 = 15 Now take the square root of both sides: X 5 = ± 15 X = _ 5 ± 15 15, 15 X — 75 = O
13
Completing the Square Recall from factoring that a Perfect-Square Trinomial is the square of a binomial: Perfect square Trinomial Binomial Square The square of half of the coefficient of x equals the constant term: = 16 [1/2 = 9
14
Completing the Square Write the equation in the form + bx = c Add to each side of the equation Factor the perfect-square trinomial + + [1/2(b)] 2 Take the square root of both sides of the equation Solve for x
15
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 Add to both sides: b = ...5 = 32 = 9 6 4+9 Now take the square root of both sides
16
5 5 5 5,-3 - 5}
17
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 + (5/2)r = (3/2) Add to both sides: b = 2 = (5/4)2 5/2 = 25/16 r2 + (5/2)r + 25/1 6 = (3/2) + 25/1 6 + (5/2)r + 25/16 = 24/16 + 25/16 (r + 5/4)2 = 49/1 6 Now take the square root of both sides
18
49/16 r = - (5/4) + (7/4) = 2/4 = 1/2 and r = - (5/4) - (7/4) - — -12/4 — —-3 r = {1/2, - 3}
19
Ex: Solve 9-5 = (p- -2) Is this a quadratic equation? FOIL the RHS 3p p2 — 2P —p +2 Collect all terms 310 — 5 = p 2 — + 2 102 _ + 7 = O A-ha Quadratic Equation -5 complete the square p2 — 6p -6 2 (-3)2
20
2 2,3- 2}
21
The Quadratic Formula Consider a quadratic equation of the form ax2 + bx + c = 0 for a nonzero Completing the square b b 4a
22
The Quadratic Formula 4ac 4a2 4a b 2a Solutions to ax2 + bx + c = 0 for a nonzero are 2a b2 — 4ac 2 — 4ac
23
Ex: Use the Quadratic Formula to solve1x2 + 7x + 6 Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by —b ± b2 — 4ac 2a Identify a, b, and c in ax2 + bx + c = 0: 6 Now evaluate the quadratic formula at the identified values of a, b, and c
24
2 2 х 20) 49 — 24 2 25 = 1 and х
25
Ex: Use the Quadratic Formula to solve 10 = O Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by _ b ± b2 — 4ac 2a Identify a, b, and c in arn2 + bm + c = 0: -10 Now evaluate the quadratic formula at the identified values of a, b, and c
26
12 — 2(2) — ± 1 -k 80 4 _ ± 81 4 4 - 2 and m = - 5/2 m
27
Any questions .
Discussion
Copyright Infringement: All the contents displayed here are being uploaded by our members. If an user uploaded your copyrighted material to LearnPick without your permission, please submit a Takedown Request for removal.
Need a Tutor or Coaching Class?
Post an enquiry and get instant responses from qualified and experienced tutors.
If you have your own PowerPoint Presentations which you think can benefit others, please upload on LearnPick. For each approved PPT you will get 25 Credit Points and 25 Activity Score which will increase your profile visibility.