Question

2x^2+6x+4=0 how do I solve this

Answer 1

You could factor it, graph it, complete the square, use the quadratic formula...any of those sound familiar?

Answer 2

No I an lost can you show me how

Answer 3

Yes, which one would you like to learn tonight? :-)

Answer 4

Factoring please

Answer 5

Good choice! They're all good choices :-)

When we solve by factoring, we're trying to rearrange our equation into a multiplication where we have a bunch of things being multiplied together, with the product equalling 0. The only way you can get a result of 0 when multiplying is if 1 or more of the things being multiplied is 0, right?

Answer 6

Ok

Answer 7

So if we can rearrange our equation to look something like \[(x-a)(x-b) = 0\]we'll know that \[x-a=0\] and \[x-b=0\]or \[x = a\]\[x=b\]are our two solutions. We can see this, because if we plug either one of them in for \(x\), we get:
\[(a-a)(a-b) = 0(a-b) = 0\]or\[(b-a)(b-b) = (b-a)0 = 0\]

The trick is in figuring out how to get there from here: \[2x^2+6x+4=0\]

Answer 8

Now, the first thing we can and usually should do is to examine the polynomial and see if we can factor out anything from each of the terms. What do you think? Is there anything we can divide out from each term?

Answer 9

The 2 and 4

Answer 10

we can't factor out a 4 unless we want a fraction as a coefficient (which we don't). But 2,6,4 are all divisible by 2. What do you get if you divide out a 2 from each term?

Answer 11

1,3,2 ?

Answer 12

Right, but can you write the whole equation with that factored out, please?

Answer 13

No that is what I am not understanding

Answer 14

\[2x^2+6x+4 = 0\]\[2(x^2+3x+2) = 0\]Agreed?

Answer 15

I'm sure you could have done that, but maybe it wasn't clear to you what I was requesting. Now you know :-)

Answer 16

Now, here's a question: does that leading 2 have any impact on the values of the solutions to this equation? Why or why not?

Answer 17

Yes

Answer 18

Because you use it to factor out

Answer 19

Hmm. Can you solve \[2x= 0\]for me?

Answer 20

Then solve \[4x = 0\]That's the same as \[2(2x)=0\]why don't we get different answers, if the leading 2 matters here?

Answer 21

The real answer is that it doesn't matter, because we can divide both sides of the equation (or multiply) by the same number and not change the solutions.

Remember, the solutions are the values of \(x\) where \[(x-a) = 0 \] with \(a\) being a solution.
\[2(x-a) = 0\]has exactly the same solution as \[5(x-a) = 0\]namely, \(x = a\).

So, we can discard that factor of 2 to start. However, if we found a factor that had a variable in it, we could NOT discard the variable. For example:
\[2x^2+2x = 0\]\[2x(x+1) = 0\]We cannot go to \[(x+1) = 0\] and just toss aside the \(x\) like yesterday's smelly leftovers. We CAN discard the 2, but not the \(x\).

Does that make sense?

Answer 22

So, now we have \[2(x^2+3x+2) = 0\]Divide both sides by 2, giving us\[x^2+3x+2 = 0\]

Now, the trick to factoring is to think about how the pieces come together when you multiply:

\[(x+a)(x+b) = x(x+b) + a(x+b) = x*x + b*x + a*x + a*b\]\[=x^2+(a+b)x + ab\]
Compare that with our polynomial to be factored:
\[x^2+3x+2\]\[x^2+(a+b)x + ab\]Clearly, those are equivalent if we can find a pair of numbers \(a,b\) such that \(a+b = 3\) and \(a*b = 2\)

What are the factors of \(2\)?
1*2
2*1

do either of those pairs of numbers add up to 3? Yes, they both do.

So our factoring is\[(x+1)(x+2) = x(x+2) + 1(x+2) = x*x + 2*x + 1*x + 1*2\]\[(x+1)(x+2) = x^2+3x+2\]

Now that we know we can write \[x^2+3x+2\]as a product of factors, we set those factors equal to 0 and solve for \(x\):\[(x+1)(x+2) = 0\]\[(x+1) = 0\]\[(x+2) = 0\]\[x+1 = 0\rightarrow x = -1\]\[x+2=0\rightarrow x=-2\]So our solutions are \[x = -1,-2\]
If we plug those back into the original equation, they should both produce a true sentence:
\[2x^2+6x+4=0\]\[2(-1)^2+6(-1)+4=0\]\[2 -6+4 = 0\]\[0=0\]\[2(-2)^2+6(-2)+4 = 0\]\[8-12+4=0\]\[0=0\]Both solutions check out!

Answer 23

I think I am getting it. Can I try 1 and you check my answer to see if I am doing it right?

Answer 24

Please, I'd be happy to do so.

Answer 25

Anyone who wants their answers checked, as opposed to given to them, is my kind of student :-)

Answer 26

I'm in the middle of preparing dinner; if I don't respond instantly, be patient, I'll be back!

Answer 27

Ok the problem is 2x^2+7x+6=0

Answer 28

Good. Want to race me? :-)

Answer 29

Lol 2(-1)^2 +7(-1)+4=0

Answer 30

2-7+4=0

Answer 31

-5+4=0
X=-1

Answer 32

2-7+4 = 0? Are you sure?

Answer 33

Factoring \[2x^2+7x+6=0\]
2*6 = 12
A pair of factors of 12 that add to 7 is 3 and 4
\[2x^2 + 3x + 4x + 6 = 0\]\[(2x^2+3x) + (4x+6) = 0\]\[x(2x+3) + 2(2x+3) = 0\]\[(2x+3)(x+2) = 0\]Set each product term equal to 0 and solve for \(x\):
\[x+2 = 0\]\[x = -2\]

\[2x+3 = 0\]\[2x=-3\]\[x = -\frac{3}{2}\]

Solution check:

\[2(-2)^2 + 7(-2) +6 = 0\]\[8-14 + 6 = 0\]\[14-14=0\]\[0=0\checkmark\]

\[2(-\frac{3}{2})^2+7(-\frac{3}2) + 6 = 0\]\[2(\frac{9}{4}) -7(\frac{3}{2}) + 6 = 0\]\[\frac{9}{2} - \frac{21}{2}+6 = 0\]\[-\frac{12}{2} + 6 = 0\]\[-6+6=0\]\[0=0\checkmark\]