Question

2x squarded - 3x -5 =0

Answer 1

2x^2-3x-5=0
factor and find the two 0's

Answer 2

yes that's the question

Answer 3

okay, work this one just like the one we just did: multiply the earpiece numbers, and find a pair of numbers that multiply to that number and add to -3

Answer 4

if i'm not mistakened it would be 2 multiplied by -5 which would make it -10?

Answer 5

Good so far. Are you saying 2* -5 because those are the numbers on the equation, or because you've already figured out that 2+ -5 = -3?

Answer 6

yeah for both of em
so it would then become 2x squared + 2 - 5 - 5?

Answer 7

right you are, so long as you stick in the x's in the middle

\[2x^2 + 2x - 5x - 5\]

Answer 8

right forgot the x's
then parenthesis?
that would make it (2x squared + 2x) - (5x -5)

Answer 9

Okay, now here's a trap for the unwary: when you put parentheses around that second group, what you're really doing is writing

\[(2x^2+2x) + (-5x-5)\]

see the difference?

Answer 10

but don't you put that on the outside then just change the sign on the inside?

Answer 11

the reason for that is that \(-(5x-5)\) is shorthand for \(-1(5x-5)\)
and \[-1(5x-5) = -1*5x -1*-5 = -5x + 5\]

Answer 12

please explain i am confused

Answer 13

Okay, we don't have any question about the left hand group, it's just the \(-5x-5\) that is causing heartburn, right?

Answer 14

yup

Answer 15

there are a couple of ways we can deal with this.

1) if you have a choice, put the "middle" numbers in the order that makes this not happen.

For example, we could have written\[2x^2 -5x + 2x -5\]and we wouldn't have to worry because we'd just have \[(2x^2-5x) + (2x-5)\]

Unfortunately, sometimes you can't do that.

2) just take it for granted that you always want a + between the two groups, and write the ( ) around the whole second thing:

\[2x^2 + 2x - 5x -5\]\[(2x^2 + 2x ) + (-5x-5)\]

This will always work. The next step is the factoring, and here we'll have
\[2x(x+1) -5(x+1)\]right?

Answer 16

but at first you would have -(5x-1)
then you would put that - in the problem and it would become (5x+1) right?

Answer 17

but we could do it your way and it would come out like that i think then it would become (2x-5) (x+1)

Answer 18

sorry, got a phone call...

Answer 19

so yes, \((2x-5)(x+1)\) is the correct factoring.

Answer 20

i'm stumped

Answer 21

going back to your mistake earlier:
(2x squared + 2x) - (5x -5)

\[(2x^2 + 2x) - (5x -5)\]the problem is that the - sign in front of the parentheses means "-1", so what we really have is \[2x^2+2x-1(5x-5)\]and if you use the distributive property to expand that,
\[2x^2+2x-1*5x-1(-5) = 2x^2+2x-5x+5 = 2x^2 -3x+5\]and that is not what we started with. We can't do anything while factoring that actually changes the value of the expression.

Answer 22

oh

Answer 23

That one little sign change transforms it into an expression that cannot be factored, so that's not good!

Answer 24

So, the safest thing (and what I personally do, even after having done hundreds of these problems) is to just always put a + in the middle, and wrap the parentheses around the - signs :

\[2x^2 + 2x -5x -5 \]\[(2x^2+2x) + (-5x -5)\]\[2x(x+1) + (-5)(x+1)\]\[2x(x+1)-5(x+1)\]\[(2x-5)(x+1)\]
any questions about that?

Answer 25

nope

Answer 26

i'm stumped on the (2x-5) and how you make it smaller where it's the x=

Answer 27

oh, \[(2x-5) (x+1) = 0\]\[(x+1) = 0\]\[(2x-5) = 0\]those last two lines come from the zero product principle, which says that if \(a*b =0\) then either \(a=0\) or \(b=0\) or both.

Is your question how to solve \[2x-5=0\]?

Answer 28

yeah

Answer 29

Okay, put on your helmet, you're going to want to hit your forehead on something hard in a second :-)

Answer 30

\[2x-5=0\]add 5 to both sides
\[2x-5+5 = 0+5\]\[2x = 5\]divide both sides by 2
\[\frac{2x}{2} = \frac{5}{2}\]\[x = \frac{5}{2} = 2\frac{1}{2} = 2.5\]

Answer 31

you've gotta be crapping me

Answer 32

personally, my preference is to keep the first one (5/2) unless you need a decimal number. the middle one is just a PITA

Answer 33

Like I said :-)

Answer 34

man i'm gonna start calling you doc cause your a life saver

Answer 35

it would end up being x= 2.5 and x= -1 correct?

Answer 36

can you help me with some more on here? it's just like four more

Answer 37

let's check them out:

\[2x^2-3x-5=0\]\[2(2.5)^2-3(2.5)-5=0\]\[2*6.25 - 7.5-5 = 0\]\[12.5-12.5=0\]
\[2(-1)^2-3(-1)-5=0\]\[2+3-5=0\]Yep, good solutions!

Answer 38

I can help you some more, but I also have to make my dinner. Can you take a shot at them and I'll check your answers? Also, if you want to just quickly post all 4, I can have a look and see if there are any traps waiting for you...

Answer 39

got you x squared = 3x + 40 it would end up being which would make it have to be multiplied by 40... then you would find a multiple of 40 that adds to 3... erm would that be....i'm stumped