Quadratic Equations by Example -- Section 1.3 --

ax2 + bx + c = 0

 Solve for x:   6x2 + 3x = 0        Factoring Method
 Solve Step Check 6x2 + 3x = 0 Factor, if possible. If x = 0, then 6(0)2 + 3(0) = 0      0 + 0 = 0 3x(2x + 1) = 0 Solve each linear equation separately 3x = 0 or 2x + 1 = 0 Zero Property of Multiplication 3x = 0   Þ   x = 0/3 = 0 If ab = 0 then a = 0 or b = 0 2x + 1 = 0   Þ    2x = -1 If x = -1/2, then 6(-1/2)2 + 3(-1/2) = 0       6(1/4) – 3/2 = 0  3/2 – 3/2 = 0 x = -1/2 {-1/2, 0} Solution Set Notation
 Solve for x:   8x2 + 2x = 0 [Solution]
 Solve for x:   2x2 + x = 6     Factoring Method
 Solve Step Check 2x2 + x – 6 =  0 Factor, if possible. If x = -2, then 2(-2)2 + (-2) = 6      8 – 2 = 6 (2x – 3)(x + 2) = 0 You must equation equal to 0 first. 2x – 3 = 0 or x + 2 = 0 Zero Property of Multiplication 2x – 3 = 0   Þ    2x = 3 If ab = 0 then a = 0 or b = 0 If x = 3/2, then 2(3/2)2 + (3/2) = 6       2(9/4) + 3/2 = 6  9/2 + 3/2 = 6 12/2 = 6 x = 3/2 Solve each linear equation separately x + 2 = 0   Þ   x = -2 {-2, 3/2} Solution Set Notation
 Solve for x:  3x2 + x = 2 [Solution]
 Solve for x:  (x + 2)2 = 3
 Square Root Method Solve Step Check (x + 2)2 = 3 Original equation If x  = -0.268, then Take square root of both sides (-0.268 + 2)2 = 3 Subtract 2 from both sides (1.732)2 = 3 2.999824 » 3  **Rounding error (exact) There are 2 distinct answers. Approximations... If x  = -3.732, then  (-3.732 + 2)2 = 3     (-1.732)2 = 3 2.999824 » 3 **Rounding error or  (exact) x » -2 + 1.732 = -0.268 x » -2 – 1.732 = -3.732 **You may notice that the approximations did equal 3 exactly.  Recall that the square root of 3 is an irrational number and can not be expressed exactly as a decimal. You can get closer by using more decimal places.
 Solve for x:  (x – 4)2 = 5 [Solution]
 Solve for x:  (5x – 2)2 = 4
 Solve Step Check (5x – 2)2 = 4 Original equation If x  = 4/5, then 5x – 2 = ± 2 Take square root of both sides (5(4/5) – 2)2 = 4 5x – 2 = 2  or 5x – 2 = -2 Solve each equation separately (4 – 2)2 = 4 5x – 2 = 2 Solve first equation (2)2 = 4 5x = 4 x = 4/5 If x  = 0, then 5x – 2 = -2 Solve second equation (5(0) – 2)2 = 4 5x = 0 (- 2)2 = 4 x = 0/5 = 0 {4/5, 0} Solution Set Notation
 Solve for x:  (3x – 1)2 = 9 [Solution]
 Solve for x:  x2 + 10x + 3 = 0   Link - Completing the Square
 Solve Step Check x2 + 10x + 3 = 0 Original equation If x  = -0.3096, then x2 + 10x      = -3 Subtract 3 from both side (-0.3096)2 + 10(-0.3096) + 3 = 0 .095852 – 3.096 + 3 = 0 -.00015 » 0 x2 + 10x + (10/2)2  = -3 + (10/2)2 Take half of 10, square it, add it to both sides (x + 5)2  = -3 + 25 **Rounding error (x + 5)2  = 22 If x  = -9.6904, then Take square root of both sides (-9.6904)2 + 10(-9.6904) + 3 = 0 93.90385 –  96.904 + 3 = 0 Solve each equation separately -.00015 » 0 Approximations... x -5 + 4.6904  = -0.3096 x -5 – 4.6904 = -9.6904 **Rounding error **You may notice that the approximations did equal 3 exactly.  Recall that the square root of 3 is an irrational number and can not be expressed exactly as a decimal. You can get closer by using more decimal places.
 Solve for x:  x2 + 6x – 1 = 0  Link - Completing the Square [Solution]

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