Three Ways to Solve Quadratic Equations
Date: 05/01/2003 at 18:15:55 From: Bronthie Subject: How do I solve this? x^2 + 4x + 2 = 0
Date: 05/04/2003 at 23:22:46 From: Doctor Dotty Subject: Re: How do I solve this? Hi Bronthie, thanks for the question. What you have there is a quadratic equation: an equation with a single variable where the largest power is a 2. There are three main ways to solve quadratic equations: * by inspection - Note: this does not work for every quadratic * by completing the square * by using the quadratic formula Which method you end up using is up to you; everyone has his or her own preference. I suggest trying each of the methods with your question and seeing which you prefer. I will use x^2 - x = 20 as an example for all three. Before starting any of the methods, you need to use addition and subtraction to get everything on one side of the equals sign. x^2 - x = 20 x^2 - x - 20 = 0 SOLVING BY INSPECTION: (Some people know this as solving by factoring.) x^2 - x - 20 = 0 First you factorise the equation into a format like: (ax + c)(bx + d) = 0 This looks a bit complicated, but it isn't really. Multiply out the parentheses: abx^2 + bcx + adx + cd = 0 and collect the x's: abx^2 + x(bc + ad) + cd = 0 Now we have to find out what 'a', 'b', 'c', and 'd' are: abx^2 + x(bc + ad) + cd = 0 x^2 - x - 20 = 0 Let's start at the left. The coefficient (number that is multiplied by) the x^2 is 1 in our equation. So: ab = 1 The easiest pair of numbers that works for 'a' and 'b' in the above equation is when they are both 1. So we can rewrite our identity: x^2 + x(c + d) + cd = 0 Now if we look at the coefficients of x and the constant at the end, we should be able to get another two equations: Coefficient of x: -1 = c + d Constant: -20 = cd You can either solve those simultaneously, or just look at them. -5 + 4 = -1 and -5 * 4 = -20. So 'c' is -5, and 'd' is 4. Usually, you probably wouldn't do all that. When you get used to what's happening, you will probably be able to get this far in your head. When the x^2 has 1 as a coefficient, all we did was to find two numbers that multiply together to get the constant and sum to get the coefficient of x. So we have: (x - 5)(x + 4) = 0 The solutions for x can be just read from this. They are the two numbers ('c' and 'd') with the signs the other way around. The reason this works is as follows: (x - 5)(x + 4) = 0 Divide both sides by (x + 4): x - 5 = 0 So x = 5. Similarly, you can find that x can also be -4. So: x = 5, -4. COMPLETING THE SQUARE: x^2 - x - 20 = 0 First you divide everything by the x^2 coefficient. As it's 1 in this case, we don't need to bother. You complete the square by getting the equation into the format: (x + b)^2 + c = 0 Let's multiply that out to see what we're aiming at: (x + b)(x + b) + c = 0 x^2 + bx + bx + b^2 + c = 0 x^2 + 2bx + b^2 + c = 0 At the moment we have: x^2 - x - 20 = 0 Let's look at the coefficient of x: -1 = 2b -1/2 = b Now, let's look at the constants: -20 = b^2 + c We know b = -1/2 so: -20 = (-1/2)^2 + c -20 = 1/4 + c -20.25 = c So that gives us: (x + b)^2 + c = 0 (x - 0.5)^2 - 20.25 = 0 This can be solved by normal algebra from here: (x - 0.5)^2 = 20.25 x - 0.5 = sqrt(20.25) x - 0.5 = 4.5 or -4.5 x = 5 or -4 Which agrees with our result from the inspection method. Again, you will see patterns after doing this for a while, so you will be able to omit more and more steps on paper. QUADRATIC FORMULA: Here, you just put all the numbers into a nice big formula and it tells you what the answer is. To use this, you need your quadratic in the form: ax^2 + bx + c = 0 Which it is already. We have: x^2 - x - 20 = 0 So a = 1, b = -1, and c = -20. Here is the quadratic formula: - b +/- sqrt(b^2 - 4ac) x = ----------------------- 2a Put in the values of 'a', 'b' and 'c': 1 +/- sqrt(1^2 - 4 * 1 * -20) x = ----------------------------- 2 * 1 1 +/- sqrt(1 + 80) x = ------------------ 2 1 +/- sqrt(81) x = -------------- 2 1 +/- 9 x = ------- 2 1 + 9 1 - 9 x = ----- or ----- 2 2 x = 5 or -4. It always strikes me how nice it is that all these methods arrive at the same answer, but by completely different routes. Does that help? If I can help any more with this problem or any other, please write back. - Doctor Dotty, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.