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5-5 Quadratic equations

5-5 Quadratic equations. Zero – product property. Solving quadratic equations. Zero product property After factoring a trinomial you can set each binomial equal to zero and solve for your variable Example: Solve x 2 + 7x =18

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5-5 Quadratic equations

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  1. 5-5 Quadratic equations Zero – product property

  2. Solving quadratic equations • Zero product property • After factoring a trinomial you can set each binomial equal to zero and solve for your variable • Example: • Solve x2 + 7x =18 • First, put all terms on the left side of the equation and set the trinomial equal to zero • x2 + 7x – 18 = 0 • Factor the trinomial into 2 binomials • (x + 9) (x-2) = 0 • Set each binomial equal to zero and solve each part. • x + 9 = 0 x – 2 = 0 • x = -9 x = 2 • The solutions are -9 and 2.

  3. Solving by factoring square roots • Example: Solve 5x2 - 180 = 0 • Rewrite the equation so the squared term is on the left side and the constant is on the right • 5x2 = 180 • Isolate the variable • Divide by 5 to isolate the x-squared term • x2 = 180/5 • x2 = 36 • Take the square root of each side to solve for x • x2 = 36 • x =+ 6

  4. Solving quadratic equations • Check the solutions back in the original problem. Sometimes one of the solutions won’t check. If the solution doesn’t check, it is thrown out. • x2 + 7x =18 x2 + 7x =18 • x = -9 x = 2 • (-9)2 + 7(-9) =18 (2)2 + 7(2) =18 • 81 – 63 = 18 4 + 14 = 18 • 18 = 18 18 = 18 (the answers check)

  5. Try these two “different” problems: 2x2 + 4x =6 16x2 = 8x • 2x2 + 4x – 6 = 0 2( x2 + 2x – 3)= 0 or 2x2 + 4x – 6 = 0 2(x+3) (x-1) = 0 or (2x + 6) (x – 1) = 0 Divide by 2 (x + 3) (x – 1) = 0 or (2x + 6) (x – 1) = 0 x + 3 = 0; x-1 =0 or 2x+6 = 0 ; x-1 = 0 x = -3; x = 1 or 2x = -6 ; x = 1 x = -3 ; x = 1 The solutions are -3 and 1 Both methods yield the same answers Check your answers in the original equation. • 16x2 - 8x = 0 • 8x (2x – 1 ) = 0 • 8x = 0 2x – 1 = 0 • x = 0 2x = 1 • x = ½ • The solutions are 0 and ½

  6. Try these two “different” problems: • 4x2 = 25 • Divide by 4 • x2 = 25/4 • x = + 5/2 • Or • 4x2 – 25 = 0 (rewrite the problem as the diff. of 2 squares) • (2x-5) (2x+5) = 0 • 2x-5 = 0 and 2x+5 = 0 • x=5/2 x = -5/2 The solutions are 5/2 and -5/2 Both methods yield the same answers Check your answers in the original equation. 4x2 - 25 =0 3x2 – 24=0 • 4x2 = 25 • Divide by 4 • x2 = 25/4 • x = + 5/2 • Or • 4x2 – 25 = 0 (rewrite the problem as the diff. of 2 squares) • (2x-5) (2x+5) = 0 • 2x-5 = 0 and 2x+5 = 0 • x=5/2 x = -5/2 The solutions are 5/2 and -5/2 Both methods yield the same answers Check your answers in the original equation. • 3x2 = 24 • Divide by 3 • x2 = 8 • x = + 8 • The solutions are 8 and - 8

  7. Homework Chapter 5 packet ; 5-5 w/s 1, 5, 8, 9, 18, 28, 30

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