pls help: Prove (2tanx-1)(tanx+2)=4tan^2x+7tanx-2

[minisue1]

I don't know how to get the solution...the closest I come is:
2tanx(tanx +2)-(tanx+2)= 2tan^2x+3tanx-2

Where am I going wrong?

 

[stapel]

I don't know how to get the solution...the closest I come is:
2tanx(tanx +2)-(tanx+2)= 2tan^2x+3tanx-2

Where am I going wrong?

Until you show your work, I'm afraid there is no way for us to help you find where you might be going wrong. Sorry!

 

[minisue1]

I have shown my work, with the expansion of 2tanx(tanx + 2)- 1(tanx + 2)= 2tan^2(x)+3tanx-2......

 

[srmichael]

I don't know how to get the solution...the closest I come is:
2tanx(tanx +2)-(tanx+2)= 2tan^2x+3tanx-2

Where am I going wrong?

Hint: FOIL the left side of the equation and then combined like terms making the equation equal to 0. You should be able to readily see what to do once you get to that point.

 

[minisue1]

I have FOIL-ed, which is how I get the answer: 2tan^2x+3tanx-2.

But, I need to prove that it =4tan^2x+7tanx-2, and that is what's baffling me.

 

[srmichael]

I have FOIL-ed, which is how I get the answer: 2tan^2x+3tanx-2.

But, I need to prove that it =4tan^2x+7tanx-2, and that is what's baffling me.

Are you sure the problem is not asking you to solve for the angle x?

What exactly does the problem state?

 

[lookagain]

I have FOIL-ed, which is how I get the answer: 2tan^2x+3tanx-2.

But, I need to prove that it =4tan^2x+7tanx-2, and that is what's baffling me.

minisue1,

no one can prove it, because it's **false**!


Look at a counterexample where \(\displaystyle \ x = 45 \ degrees: \ \)


\(\displaystyle [2tan(45 \ degrees) \ - \ 1][tan(45 \ degrees) \ + \ 2] \ ** = \)

[2(1) - 1][(1) + 2] =

(1)(3) =

3




But \(\displaystyle 4tan^2(45 \ degrees) \ + \ 7tan(45 \ degrees) - 2 \ =\)

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

4 + 7 - 2 =

9


- - - - - - - - - -




** I don't know why the two close brackets look faint (at least to me).

 

[minisue1]

The first part of the problem says:

Show that (2tanx-1)(tanx+2) = 4 tan^2x + 7 tanx - 2.

Then part ii) says Hence, or otherwise, solve 4tan^2x + 7tanx -2 for all x in 0≤x<∏/2

I have literally wasted hours it, and it is frustrating me!!!

 

[soroban]

Hello, minisue1!

Are you sure you didn't copy the problem incorrectly?

\(\displaystyle \text{(a) Show that: }\: ({\color{red}4}\tan x-1)(\tan x+2) \:=\: 4\tan^2x + 7\tan x - 2\)

\(\displaystyle \text{(b) Hence, or otherwise, solve: }\:4\tan^2x + 7\tan x -2\:\color{red}{=\:0}\:\text{ for }0 \le x \le \frac{\pi}{2}\)

 

[minisue1]

No I have copied it 100%! It must be a typo on the paper, as I have completed Part 2, which DID give me (4tanx-1)(tanx+2)....What a waste, I spent hours going over it!!!

I thought I was doing something wrong!