1. ## Algebra problems..

Will someone show me the steps to solving these problems. I have finals this monday and I'm not ready yet *.

x^2 + 7x + 6 = 0

6n^2 + 7n = 20

and

y^2 - 3y/5 + 2/25 = 0

2. Originally Posted by almarie15j
Will someone show me the steps to solving these problems. I have finals this monday and I'm not ready yet *.

x^2 + 7x + 6 = 0
$\displaystyle (x+1)(x+6)=0$

Make each factor zero.

3. Originally Posted by almarie15j
Will someone show me the steps to solving these problems. I have finals this monday and I'm not ready yet *.

x^2 + 7x + 6 = 0

6n^2 + 7n = 20

and

y^2 - 3y/5 + 2/25 = 0
there are three main ways to deal with these problems. i expect that you are required to know all of them for your final. which method do you need help with? do you know when to use which?

- foiling (or factoring), which is what TPH demonstrated earlier
- completing the square

4. Originally Posted by Jhevon
there are three main ways to deal with these problems. i expect that you are required to know all of them for your final. which method do you need help with? do you know when to use which?

- foiling (or factoring), which is what TPH demonstrated earlier
- completing the square
So far we've been studying about foiling and the quadratic formula, not the completing the square though. I don't think we've done that, unless I prob. forgot it which is bad considering finals is this monday :s.

But the thing is I don't know when to use which though..

5. Originally Posted by ThePerfectHacker
$\displaystyle (x+1)(x+6)=0$

Make each factor zero.
SO basicly just do

x + 1 = 0 and subtract 1 on both side.

x + 6 = 0 and subtract 6 on both side and that's the answer?

6. Originally Posted by almarie15j
SO basicly just do

x + 1 = 0 and subtract 1 on both side.

x + 6 = 0 and subtract 6 on both side and that's the answer?
Yes. Just solve for "x":

$\displaystyle x=-1$ and $\displaystyle x=-6$.

And those are the two possible answers.

7. Originally Posted by almarie15j
So far we've been studying about foiling and the quadratic formula, not the completing the square though. I don't think we've done that, unless I prob. forgot it which is bad considering finals is this monday :s.

But the thing is I don't know when to use which though..
with foiling, this is the procedure. let's use the problem TPH worked out as an example.

$\displaystyle x^2 + 7x + 6 = 0$

When we want to foil we have to come up with two numbers. these numbers must be such that when we multiply them together we get the constant term (we must account for the sign), and when we add them together we get the coefficient of $\displaystyle x$ (we must also account for the sign). So as you see, here we must come up with two numbers that when multiplied we get +6 and when added, we get +7. it is simple enough to come up with such numbers, and since TPH did the problem already, it would be no surprise that these numbers are +1 and +6. since (+1)*(+6) = +6 our lone constant, and +1 + 6 = + 7 our coefficient of $\displaystyle x$

once we have these two numbers, we simply foil the expression putting them in two sets of brackets with $\displaystyle x$'s in front of them. so we get:

$\displaystyle (x + 1)(x + 6) = 0$

now if two numbers when multiplied gives a result of zero, it means one or the other is zero. so then:

either $\displaystyle x + 1 = 0$ or $\displaystyle x + 6 = 0$

then $\displaystyle x = -1$ or $\displaystyle x = -6$

How do we know when to use the above method and when to use the quadratic formula? Simple. If you can't come up with two numbers, use the formula.

Now to refresh your memory, the quadratic formula is given and explained below.

Given a quadratic of the form: $\displaystyle ax^2 + bx + c = 0$

The roots of the quadratic (that is the $\displaystyle x$-values that make the quadratic = 0) are given by the formula:

$\displaystyle x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

Now, try the others

8. Originally Posted by almarie15j

But the thing is I don't know when to use which though..
So we have 3 methods:

completing the square, quadratic formula, factoring

You want to try to use factoring first. It's much easier. If you have a quadratic ($\displaystyle ax^2\,+\,bx\,+\,c$) with an $\displaystyle a$ of 1, you want to look to use factoring. If it don't work, use complete the square/quadratic formula. If $\displaystyle a$ is any real number besides 1 or 0, you want to look to use the "ac" method, which is a variation of factoring. If it don't work, use the quadratic formula/complete the square.

Do you all agree with this?

9. Originally Posted by Jonboy
So we have 3 methods:

completing the square, quadratic formula, factoring

You want to try to use factoring first. It's much easier. If you have a quadratic ($\displaystyle ax^2\,+\,bx\,+\,c$) with an $\displaystyle a$ of 1, you want to look to use factoring. If it don't work, use complete the square/quadratic formula. If $\displaystyle a$ is a 2, you want to look to use the "ac" method, which is a variation of factoring. If it don't work, use the quadratic formula/complete the square.

Do you all agree with this?
sounds ok, but you may want to mention what the "ac method" is . which i would have done when almarie15j did the other problems (assuming you are talking about what i think you are talking about). and you should have said, if $\displaystyle a$ is any real number other than 1 or 0 use the ac method or complete the square or the formula (and if you can't factor when a = 1).

This is my 21th post!!!!!!!

10. Originally Posted by Jhevon
with foiling, this is the procedure. let's use the problem TPH worked out as an example.

$\displaystyle x^2 + 7x + 6 = 0$

When we want to foil we have to come up with two numbers. these numbers must be such that when we multiply them together we get the constant term (we must account for the sign), and when we add them together we get the coefficient of $\displaystyle x$ (we must also account for the sign). So as you see, here we must come up with two numbers that when multiplied we get +6 and when added, we get +7. it is simple enough to come up with such numbers, and since TPH did the problem already, it would be no surprise that these numbers are +1 and +6. since (+1)*(+6) = +6 our lone constant, and +1 + 6 = + 7 our coefficient of $\displaystyle x$

once we have these two numbers, we simply foil the expression putting them in two sets of brackets with $\displaystyle x$'s in front of them. so we get:

$\displaystyle (x + 1)(x + 6) = 0$

now if two numbers when multiplied gives a result of zero, it means one or the other is zero. so then:

either $\displaystyle x + 1 = 0$ or $\displaystyle x + 6 = 0$

then $\displaystyle x = -1$ or $\displaystyle x = -6$

How do we know when to use the above method and when to use the quadratic formula? Simple. If you can't come up with two numbers, use the formula.

Now to refresh your memory, the quadratic formula is given and explained below.

Given a quadratic of the form: $\displaystyle ax^2 + bx + c = 0$

The roots of the quadratic (that is the $\displaystyle x$-values that make the quadratic = 0) are given by the formula:

$\displaystyle x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

Now, try the others
Oh yeah I remember quite remember it now. I thought this kind of problems would be solved differently and not the factoring solving kind.

So if I were to solve this problem 6n^2 + 7n = 20 would it be

6n^2 + 7n + 20 and then multiply 6 & 20 and then find the 2 numbers of 120 that will add to positive seven. Then once I find the 2 numbers, do the grouping part. I think I might be confusing you? Thanks for the help with the other problem

11. Originally Posted by almarie15j
Oh yeah I remember quite remember it now. I thought this kind of problems would be solved differently and not the factoring solving kind.

So if I were to solve this problem 6n^2 + 7n = 20 would it be

6n^2 + 7n + 20 and then multiply 6 & 20 and then find the 2 numbers of 120 that will add to positive seven. Then once I find the 2 numbers, do the grouping part. I think I might be confusing you? Thanks for the help with the other problem
you didn't confuse me, and yes you are correct. you would multiply 6 and 20 and get 120, and then try to think of two numbers that when multiplied gives +120 and when added gives +7. you would then split the middle term (the 7x) into these two numbers, and factor by groups. of course, if you can't think of two such numbers relatively quickly on an exam, you would just use the quadratic formula, which i gave you above.

Originally Posted by almarie15j

y^2 - 3y/5 + 2/25 = 0
multiply through by 25 and then tell me where you think you should go from there