MATHICK! - Interactive Math Learning

How to solve:
Solve for x: 3x² + 17x + 10 = 0

(ax + b)(cx + d) =
(ac)x² + (ad + bc)x + bd=
3x² + 17x + 10

Your goal is to find numbers, a, b, c, and d so that

ac = 3, ad + bc = 17 and bd = 10.

Here are the possibilities for a and c that make ac = 3... a = 1, c = 3 a = 3, c = 1 a = -1, c = -3 a = -3, c = -1

and here are the possibilities for b and d that make bd = 12: b = 1, d = 10 b = -1, d = -10 b = 2, d = 5 b = -2, d = -5 b = 5, d = 2 b = -5, d = -2 b = 10, d = 1 b = -10, d = -1

Now find the combination of these that makes ad + bc = 17

Try out a problem here to get a feel for doing this. a = 3, c = 1 b = 2, d = 5

Factorization: (3x + 2) (x + 5) = 0.

Set each factor equal to 0: 3x + 2 = 0, x + 5 = 0.

Solve each equation: x = -2/3, x = -5.

Here's one to try:


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Here are the possibilities for a and c that make ac = 3...	`a = 1, c = 3 a = 3, c = 1 a = -1, c = -3 a = -3, c = -1`
and here are the possibilities for b and d that make bd = 12:	`b = 1, d = 10 b = -1, d = -10 b = 2, d = 5 b = -2, d = -5 b = 5, d = 2 b = -5, d = -2 b = 10, d = 1 b = -10, d = -1`
Now find the combination of these that makes ad + bc = 17 Try out a problem here to get a feel for doing this.	`a = 3, c = 1 b = 2, d = 5`
Factorization:	`(3x + 2) (x + 5) = 0.`
Set each factor equal to 0:	`3x + 2 = 0, x + 5 = 0.`
Solve each equation:	`x = -2/3, x = -5.`