Solving Quadratic EquationsDate: 2/15/96 at 15:4:3 From: Llano Net Internet Software Subject: quadratic equations I am very confused on how to do these problems. Please help me. My first two questions are: (3x-1) to the second power = 1 divided by 4 Instructions: Solve each equation by using the square root property. (x-3) to the second power = -20 This has the same instructions. _________________________________________ My second question is 2x to the second power - 5x + 2 = 0 Instructions: Solve each equation by factoring. _________________________________________ My next question is: (2+ -3)(2 + 1) = 5 Same instructions ______________________ My last question is 2x to the second power + 3x - 2 = 0 Instructions: Solve each equation using the quadratic formula. Date: 4/29/96 at 10:45:35 From: Doctor Steven Subject: Re: quadratic equations #1) (3x-1)^2 = 1/4 then (3x-1) = Sqrt(1/4) = 1/2 or -1/2 so 3x-1 = 1/2 or -1/2. Add one to both sides and get: 3x = 3/2 or 1/2 Divide by three on both sides to get: x = 1/2 or 1/6 #2) (x-3)^2 = -20 then x-3 = Sqrt(-20) = 2Sqrt(-5) or -2Sqrt(-5). Add three to both sides and get: x = 3 + 2Sqrt(-5) or 3 - 2Sqrt(-5). #3) 2x^2 - 5x + 2 = 0. Then we know we have two factors that look something like (?x + ?)* (?x + ?). We can figure out what the question marks are by looking at the constants in front of the x's. Note the constant at the end of the equation is 2. How can we get 2 by multiplying two numbers together. The answers are 2*1 and -2*-1. So our possibilities are: (?x + 2)*(?x + 1) or (?x - 2)*(?x - 1). We want the middle number to be negative so we can rule out the first choice, so we know (?x - 2)*(?x - 1). We also know we want the x^2 term to be multiplied by 2. How can we get this? Well again the answers are 2*1 or -2*-1. Since we try to keep negatives out of our highest term we throw the negatives out and say it must be 2*1. So now the question is where does the 2 go and where does the 1 go. We can check to see which one it is: (2x -2)*(x - 1) = 2x^2 - 3x + 2, (x - 2)*(2x - 1) = 2x^2 - 5x + 2. So the factorization of 2x^2 - 5x + 2 = 0 is (x - 2)*(2x - 1) = 0. The only way for a product to be zero is if one of the multipliers is zero, so if: 1) x - 2 = 0, or 2) 2x - 1 = 0, then their product will be zero (i.e, 2x^2 - 5x + 2 will be zero). So the solutions to this are x = 2, and x = 1/2. #4) In this one I'm assuming you meant (2x + -3)(2x + 1) = 5. We get by using FOIL on the left side: 4x^2 - 4x - 3 = 5. Subtracting 5 from both sides we get: 4x^2 - 4x - 8 = 0. Dividing by four we get: x^2 - x - 2 = 0 Factor this using the method above and obtain: (x - 2)*(x + 1) = 0. So the solutions are: x - 2 = 0, and x + 1 = 0. So x = 2, or -1. #5) 2x^2 + 3x - 2 = 0. The quadratic equation states that for a quadratic equation of the form: ax^2 + bx + c = 0 its solutions are: x = [-b + Sqrt( b^2 - 4ac )]/2a and x = [-b - Sqrt( b^2 - 4ac )]/2a. In this case a = 2, b = 3, c = -2. Plug them in and get the solutions as: x = [-3 + Sqrt( 9 - 4*2*-2 )]/4 x = [-3 - Sqrt( 9 - 4*2*-2 )/4 Simplify to get: x = [-3 + Sqrt(25)]/4 x = [-3 - Sqrt(25)]/4. Simplify some more to get: x = 2/4 = 1/2 x = -8/4 = -2. So the roots of this are x = 1/2, and x = -2. Hope this helps. -Doctor Steven, The Math Forum |
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