Equations?

Question

Can someone tell me HOW to solve these equations? I'm having some trouble with them. Thanks!

1. x^2 + 3x - 4 =0
2. 6x^2 + 5x + 1 = 0
3. x^2 - 36 = 0
4. x^2 + 5x + 6 = 0

Answer 1

All you are doing is factoring here.
1. x^2+3x-4
you have to be able to times two #'s together to equal the far right # while adding the same two #'s to equal the middle number
(x-4)(x+1)
set x equal to zero
making x=4 and -1
2.
Step by Step method for factoring Ax2 + Bx + C

*

Step 1. Multiply together AC and list the factors of AC.
Step 2. Find a pair that adds to B. If you cannot find such a pair then the trinomial does not factor.
Step 3. Rewrite the middle term as a sum of terms whose coefficients are the chosen pair.
Step 4. Factor by grouping.
so AC=6
so you can either have (1,6)or (2,3)
(6x^2+2x)(3x+1)
2x(3x+1) 1(3x+1)
(3x+1)(2x+1)
x= -(1/3) and -(1/2)

3. 6 times 6 is 36
you can only do this when it is x^2- a number
so it is (x-6)(x+6)
4. (x+3)(x+2)
x= -3 and -2

Answer 2

All of these problems involve factoring. I'll do #2 and #3 and you can figure out the rest.

2. 6x^2 + 5x + 1 = 0

Factor the quadratic.

(2x + 1) (3x + 1) = 0

Equate each factor to 0, to get all solutions.
i) 2x + 1 = 0
2x = -1
x = -1/2

ii) 3x + 1 = 0
3x = -1
x = -1/3

Therefore, x = { -1/2, -1/3 }
(There are two solutions).

3. x^2 - 36 = 0

Factor once more. This is a difference of squares.

(x - 6)(x + 6) = 0

Therefore,
x - 6 = 0
x + 6 = 0

Implying
x = { 6, -6 }

Answer 3

2) 6x^2 + 5x +1 the last term must be +1 in each factor, 1, 2, 3, 6 are the possibilities for the first.

( 3x + 1)(2x + 1)

http://www.coolmath.com

Answer 4

Factor the trinomial, then use the zero product principle:
x^2 + 3x - 4 = 0
(x - 1)(x + 4) = 0
x - 1 = 0 OR x + 4 = 0
x = 1 OR x = -4

Answer 5

1.First of all, to get x^2 you must have two x's in the parenthesis
(x )(x )
to get a negative number you must have a postive and a negative
(x+ )(x- )
Then what adds to get three and multiplies to get -4?
(x+4)(x-1)
Check using foil
F=x^2
O=-1x
I=4x
L=-4
Simplify x^2-1x+4x-4
x^2+3x-4=0

2.to get 6x^2 you can use either 6 and 1 or 3 and 2
let's try 3 and 2
(3x )(2x )
(3x+ )(2x+ )
(3x+1)(2x+1)
Check using foil

3.(x )(x )
(x- )(x+ )
(x-6)(x+6)
Check using Foil

4.(x )(x )
(x+ )(x+ )
(x+2)(x+3)
Check using foil

I hope this helps
Email me if you have any questions

Answer 6

Alternatively, use the quadratic formula.

If you can quickly see how to factor the expression, then factoring is the way to go, as the answers just pop right out. But sometimes it's not obvious how to do the factoring. Some days you can't factor something that would be a snap on better days.

In those cases, use our old standy, the quadratic formula. A bit messier, but it always works.