Question

im stuck, help me :( (one question done, one more left)

Answer 1

"Solve the equation 4xˆ2 -4x + 7x - 7 = 2(x-1) by whichever method seems easiest. Explain why you decided to use the method you did."

Answer 2

I'm stuck on which method I should find... none come to mind.
What I did so far is simplify the equation like this:

4xˆ2 - 4x +7x - 7 = 2(x-1)
4xˆ2 + 3x - 7 = 2x - 2
4xˆ2 + x - 5 = 0

Answer 3

After that, I don't know what to do... please help

Answer 4

(don't know if I'm in the correct path either)

Answer 5

looks ok to me

Answer 6

hey you again! : )

Answer 7

is that all though...? no further simplification?

Answer 8

you could factor it or use completing the square method

Answer 9

how do i do any of those? :s

Answer 10

you know how to complete the square factoring is just about using logic

Answer 11

Factoring Looking at the standard equation

ax^2 + bx + c
4xˆ2 + x - 5
we know
a = 4
b = 1
c = -5

so first we write

(x + )(x + )

now we know that a = 4 so we have to account for that (just a guess)

(4x + )(x + )

what time what gives c = -5 but adds or subtracts together to give b = 1

Answer 12

can you solve the rest?

Answer 13

I have a rule against teaching factoring I hate doing it, there are over 9000 questions related to factoring on this site

Answer 14

"what time what gives c = -5 but adds or subtracts together to give b = 1"

nothing... the factors of -5 are -5, -1, 1, 5

Answer 15

if i multiply with any of those, i wouldn't end up getting 1 with their addition or subtraction

Answer 16

well consider that you have a 4x in the brackets

Answer 17

and think about how things multiply out


(Lx + b)(x+c) = Lx^(2) + Lx(c) + b(x) + (c)b, that Lx is going to mess with your addition, let hero cover this though he is better at this precalc stuff

Answer 18

"Solve the problem, then explain why you did what you did". Don't ya just love those kinds of questions?

Answer 19

(4x+5) (x-1) !!!!

Answer 20

Hero, definitively :p

Answer 21

Okay, so now that you've figured out the steps, what will you say to explain why you chose the method you used?

Answer 22

(4x+5) (x-1)

4x + 5 = 0
4x = -5
x = -1.25

and,

x-1 = 0
x = 1

Answer 23

well, I put, "I decided to use the method I did because I immediately recognized that I could simplify the equation and then factor it out, finding two answers."

Answer 24

Okay, I guess that's good enough :P

Answer 25

Awesome!! :D

I got another one, now that you're here, but this one is tricky..... at least for me.

"Find a radical equation of the form (square root)ax+b = x + c so that one solution is extraneous. Show the steps in solving the equation.

a) Is there a value for h that makes it possible for the equation (Square root)x+h (end square root) + 5 = o to have any real number solutions? Explain.

b) Explain the relationship between the solutions to the equation (square root)x-3(end square root)-2 = 0 and the graph of the function (square root)x-3(end square root)-2 = 0."

Answer 26

In the first part it is just asking I'm pretty sure for you to find an equation where one of the solutions is real and the other is imaginary

Answer 27

remember!!! you cannot take the square root of a negative

Answer 28

i know we have covered that (i remember imaginary numbers), but i complete forgot this :\

Answer 29

well can you at least make an attempt?

Answer 30

okay lets see so first i would have to substitute the variables with numbers, and include a negative number in the square root to make it an imaginary?

Answer 31

i meant, to have an imaginary solution

Answer 32

Try

\(\large\sqrt{2x + 1} = x - 1\) and see what happens.

Answer 33

thank you! I'm so sorry I feel alseep yesterday....... :\