how do you get the answer? its Algebra 1?

Question

x(4x+7)=0

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

(2x-1)(x+4)=0

x^2-7x+10=0

don't you divid something? Idk. pleae help I have a quiz tomorrow

THANKYOU

Answer 1

The first three are all similar.

As you may recall, if you have a*b = 0, then either a = 0 or b = 0 (or both).

For your first question, you have:
x(4x + 7) = 0
Look at the product. You have x times (4x + 7). So either x = 0 or (4x + 7) = 0.
Well, x = 0. That's simple enough. You must solve for x when (4x + 7) = 0. This is true when x = -7/4.
So, x = 0 OR x = -7/4.

The next two equations are similar. You have
3x + 2 = 0 OR
2x + 2 = 0
When you solve for this, you see that:
x = -2/3 OR x = -1

And you have
2x - 1 = 0 OR
x + 4 = 0
So, you solve both of them and get:
x = 1/2 OR x = -4

The fourth one requires an extra step. You need to factor this equation. This takes some practice. When you have an equation in the form of:
ax^2 + bx + c = 0
then you can usually factor it (there is a nifty equation you can use for those times that factoring is not possible, but I don't think you're at that stage yet).

Factoring is an art form. You just look at the factors of a and c. In this case, a = 1, so you have a = 1*1. Since c = 10, you could have c = 1*10 or c = 2*5. You usually pair up the factors of a with the factors of c. You can see that 2*1 + 5*1 = 7. Your middle term is -7, so you're on the right track.

Since the c term is positive, you know that you have the same sign. You can see that:
x^2 - 7x + 10 = (x - 2)(x - 5) = 0

If you don't know how I got to that stage, check your book again. It's hard to explain in words. Now that you have that factored, you can apply the same thing as the first three. You see that:
x = 2 OR x = 5

Hope that helps.

Answer 2

x(4x+7)=0 implies that x=0 or 4x+7=0
solving for x: 4x= -7, therefore, x= -7/4 and x=0
the solution set x ={0, -7/4}

3x+2=0 and/or 2x+2=0
x=-2/3 and x=-1

the final problem requires factoring
Initially, find two numbers whose product is positive 10 and whose sum is -7
negative 5 and negative 2 fulfill our requirements
x^2-7x+10=0
(x-5)(x-2)=0
proceed as formally described
x-5=0
x-2=0
x={5,2}

Answer 3

1) x(4x+7)-->x=0 or 4x+7=0
x=0 or 4x=-7
x=0 or x =-7/4
2) (3x+2)(2x+2)=0 -->3x+2=0 or 2x+2 =0
3x=-2 or 2x=-2
x=-2/3 or x=-1.
3)x^2 - 7x +10 = 0 --->(x-2)(x-5) --->x-2=0 or x-5=0
x=2 or x=5.

Answer 4

x(4x + 7) = 0
when
x = 0
and
4x + 7 = 0
Solve for x

The next two work the same way. The last one you have to factor, and set each factor equal to 0.

Answer 5

the answers are
1) x=0, -7/4
2) x= -2/3, 1
3) x= 1/2, 4
4) x= 2, 5
they all have two answers

Answer 6

answers algebra 1 mcdougal littell

Answer 7

You have to mulitply the problems out then add the like numbers, then divide across.
x(4x+7)=0
4x*x+7x=0

Then I get stuck. Sorry.

Answer 8

read your text
THANKYOU