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Quarativ Equations

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PPT on quadratic equations

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    QUADRATIC EQUATIONS MSJC San Jacinto Campus Math Center Workshop Series Janice Levasseur
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    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.
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    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}
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    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
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    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 [email protected] Solution: x 8+5-0 = _ 5 ...+ x = 5} repeated root
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    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)
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    ...> 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, - }
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    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
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    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
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    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
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    - 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—
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    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
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    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
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    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
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    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
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    5 5 5 5,-3 - 5}
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    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
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    49/16 r = - (5/4) + (7/4) = 2/4 = 1/2 and r = - (5/4) - (7/4) - — -12/4 — —-3 r = {1/2, - 3}
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    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
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    2 2,3- 2}
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    The Quadratic Formula Consider a quadratic equation of the form ax2 + bx + c = 0 for a nonzero Completing the square b b 4a
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    The Quadratic Formula 4ac 4a2 4a b 2a Solutions to ax2 + bx + c = 0 for a nonzero are 2a b2 — 4ac 2 — 4ac
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    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
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    2 2 х 20) 49 — 24 2 25 = 1 and х
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    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
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    12 — 2(2) — ± 1 -k 80 4 _ ± 81 4 4 - 2 and m = - 5/2 m
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    Any questions .

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