Equation RootsDate: 01/22/2001 at 19:41:18 From: Sara Robichaux Subject: Re: Algebra II Dr. Math, We are working on quadratic formulas, where we find the discriminant. Here's what I've done, and where I get stuck: 3xsquared + 5x - 2 = 0 a = 3, b = 5, c = -2 bsquared - 4ac = (5)squared - 4(3)(-2) = 25 + 24 = 49 Now I get stuck. How do you get the real, imaginary, rational, and irrational(ROOTS) of the equation? -Sara Date: 01/23/2001 at 21:19:00 From: Doctor Ian Subject: Re: Algebra II Hi Sarah, You're right on track, but I'll walk you through the whole problem. The original equation is 3x^2 + 5x - 2 = 0 This is a quadratic equation, so it has two roots. These are: -b + sqrt(b^2 - 4ac) x = -------------------- 2a and -b - sqrt(b^2 - 4ac) x = -------------------- 2a You've correctly noted that a = 3, b = 5, and c = -2. So let's substitute those into the equation: -b +/- sqrt(b^2 - 4ac) x = ---------------------- 2a -(5) +/- sqrt((5)^2 - 4(3)(-2)) x = ------------------------------- 2(3) -5 +/- sqrt(25 - (-24)) x = ----------------------- 6 -5 +/- sqrt(49) x = --------------- 6 -5 + 7 -5 - 7 x = ------- or ------ 6 6 x = 2/6 or -12/6 x = 1/3 or -2 And these are the roots. You can check them by substituting them back into the original equation: 3(1/3)^2 + 5(1/3) - 2 = 0 3(1/9) + 5/3 - 2 = 0 1/3 + 5/3 - 2 = 0 Looks good. 3(-2)^2 + 5(-2) - 2 = 0 3(4) - 5(2) - 2 = 0 12 - 10 - 2 = 0 Looks good. Note that these are both real roots, so we _could_ have factored the original equation this way: 3x^2 + 5x - 2 = (3x - 1)(x + 2) But factoring is trickier when the first coefficient (a) is something other than 1... As I said, you were already on track. Are you able to follow what I've done here? Is there any part of it that you don't understand? Would you like to move on to another problem? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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