Can you teach me how to solve quadratic equations?

Question

I've heard there are two ways of solving quadratic equations. One way is to use the quadratic formula and the other way is to factor. I've been leaning toward factoring because I thought it would be easier, but I'm still having trouble. Can someone tell me a good way to factor? Or should I just forget factoring and study the quadratic formula?

Answer 1

Using an equation is the lazy, mechanical way to do it. If you want to understand what's going on, factor it if possible.

If you *really* want to know what's going on, derive the quadratic formula yourself. It isn't as hard as it looks, and it can be done on one page.

Answer 2

The general form of a quadratic equation is ax² + bx + c = 0, where a ≠ 0. When a = 1, this makes the equation x² + bx + c = 0. When in this form (when a = 1), the procedure is as follows:
x² + bx + c = (x + m)(x + n), where (m)(n) = c and m + n = b.
Example:
x² - 7x + 12 --- We are looking for a 2 numbers. We can start by listing (or thinking) of the factor pairs of 12 (which is "c").
1 and 12 (or -1 and -12)
2 and 6 (or -2 and -6)
3 and 4 (or -3 and -4)
Remember, we want the pair that, when added together, equal -7 ( which is "b"). As you can see, in this example, -3 and -4 do the trick. So using our template from above, m = -3 and n = -4, mn = 12, and m + n = -7. Plug our m and n into the form of the procedure:
(x + m)(x + n)
(x + (-3))(x + (-4)) or (x - 3)(x - 4)
Try using these steps to factor this expression: x² + 9x + 20 ...

**Note: Suppose x² + bx + c = (x + m)(x + n). We know certain facts about m and n. (1) m and n have the same sign if is "c" is positive. They are positive if b is positive, and they are negative if b is negative. (2) m and n have opposite signs if c is negative. The larger number is positive if b is positive and negative if b is negative.

Now, the forn ax² + bx + c = 0 is only different if a is something other than 1 (or 0, but if a = 0, then it isn't a quadratic equation).I solve these problems by the "grouping method".

~ Grouping Method for Factoring Trinomials of the Form ax² + bx + c ~
(1) Obtain the grouping number (a)(c).
(2) Find the factor pair of the grouping number whose sum is (b).
(3) Use those two factors to write (bx) as a sum of two terms.
(4) Factor by grouping.
Example: Factor 2x² + 19x + 24.
Solution:
(1) The grouping number is (a)(c) = (2)(24) = 48.
(2) The factor pairs of 48 are: 48*1, 24*2, 16*3, 12*4, and 8*6. Now b = 19 so we want the factor pair of 48 whose sum is 19. Therefore, we select the factors 16 and 3.
(3) We use the numbers 16 and 3 to write 19x as the sum of 16x and 3x.
2x² + 19x + 24 = 2x² + 16x + 3x + 24
(4) Factor by grouping.
2x² + 16x + 3x + 24 = 2x(x + 8) + 3(x + 8)
= (x + 8)(2x + 3)
Try using these steps to factor the following equation: 3x² + 2x - 8.

Now that you have your factors, you can solve:
ax² + bx + c = 0
2x² + 19x + 24 = 0
(x + 8)(2x + 3) = 0 --- This means that either (x + 8) or (2x + 3), or both, must equal 0. Solve both:
x + 8 = 0
x = -8
2x + 3 = 0
2x = -3
x = -(3/2)
So your answer would be x = {-8, -(3/2)}

If none of these ways works, then use the quadratic formula. Sometimes, it's nearly the only way to solve it...

Good luck! I hope this helped you...

Answer 3

Am I glad you asked. I've been trying to get people to factor rather than use the quadratic formula.

When you have x^ + bx +c = 0

see if you can find two numbers that when multiplied equals c and when added equal b. You frequently can. If not, then use the quadratic formula.

Consider this example:

x^2 + x -72 = 0

Can you think of two factors of 72 that add to +1 ? (psst, how about 9 and -8)...

Answer 4

I can't teach you how to solve quadratic equations, but as a Calculus-II student, I can give you this advice:

Just learn the quadratic formula.
It is SO easy to use, once you know it!

In order to memorize it, just use all the rote-memorization tricks that have worked for you in the past - perhaps writing it over and over again (worked for me), or keeping it on a card that you bring with you everywhere, and glance at it (except during quizzes and exams!).

Trust me - it looks scary and complicated, but as long as you don't have to know how it was derived (I learned, and really it's not that hard, but it's easier not to know), you'll be equipped to solve any quadratic equation!

Good luck!

Answer 5

y=Ax^2+Bx+C. For factoring, look for two numbers that multiply to AC and add up to B.