Question

Where does the graph of y = 2x2 − 9x − 5 cross the x-axis?

(−1 over 2, 0) and (5, 0)

(1 over 2, 0) and (−5, 0)

(−5 over 2, 0) and (1, 0)

(5 over 2, 0) and (−1, 0)

Answer 1

\[\Large 2x^2 − 9x − 5=0\]
can you factor this...?

Answer 2

no?

Answer 3

Do you know how to factor something like x^2 + 3x + 2 into (x+2)(x+1)?

Answer 4

yes

Answer 5

Okay, well we can use a similar method here... but you have to change it a bit.
\[\Large 2x^2 − 9x − 5=0\]

Start by multiplying the first number by the last number: 2*(-5) = -10

now you need to find two numbers that multiply to -10, and add up to the middle number, -9...

Answer 6

10 & -1?

Answer 7

Good... but it should be -10 and +1, since they add up to -9

now you need to break up that -9x into -10x and +1x:
\[\Large 2x^2 − 10x + x − 5=0\] I'll put thebrackets here for the next step
\[\Large ( 2x^2 − 10x) + (x − 5)=0\]
now you can factor the 2x out of the first brackets, \[\Large 2x(x − 5) + 1(x − 5)=0\] and one more step...

Answer 8

Can you take out the common factor?

Answer 9

you lost me

Answer 10

which step...?

Answer 11

last one

Answer 12

\[\Large ( 2x^2 − 10x) + (x − 5)=0\]
notice you can factor a 2x out of the first brackets? it's a common factor of bother 2x^2 and -10x

Answer 13

ok so how do i get the final answer

Answer 14

What common factor do you see here?\[ \Large 2x(x − 5) + 1(x − 5)=0 \]
they both have (x-5), so pull that out the front

Answer 15

like so

(x-5)(2x+1) = 0

now you can easily find the solutions.

Answer 16

5, -1/2?

Answer 17

but theres two answers?

Answer 18

oh wait, nvm, its A?

Answer 19

Good job :)

Answer 20

could you help me with one more?

Answer 21

Sure

Answer 22

Identify the solutions of 2x2 + 13x + 6 = 0.

x = 1 over 2, x = 6

x = −1 over 2, x = −6

x = −3 over 2, x = −2

x = 3 over 2, x = 2

Answer 23

2x^2 + 13x + 6 = 0

do it the same way. Multiply first by last: 2*6 = 12

now find two factors of 12 that add up to the middle 13.

Answer 24

12 & 1

Answer 25

so 2x(x + 12)( x+1) ?

Answer 26

2x^2 + 12x + 1x + 6 = 0

(2x^2 + 12x) + (1x + 6) = 0

so factor a 2x out of the first brackets
2x(x+6) + 1(x+6) = 0

Answer 27

Then just take out the common factor (x+6)

Answer 28

x = 6

Answer 29

and the other one idk

Answer 30

2x(x+6) <- dont get how to do that

Answer 31

hold on, first go from this
2x (x+6) + 1 (x+6) = 0
take out the (x+6) to get:
(x+6) (2x+1) = 0
now find your x's. 6 isn't one.

Answer 32

-6 , -1/2 ?

Answer 33

Good :D

Answer 34

thanks!

Answer 35

Welcome :)