C3 - How do I solve this equation?

I know the answer is not a lucky coincidence, i know my method is correct according to me, but Hasufel disagrees, so i feel a better mathematician should explain him.

He is right, look at the graphs.

Guys, relax! I got this!

http://www.wolframalpha.com/input/?i=sqrt+x+%3D+x%5E2+-+4x+%2B+2

Unfortunately Raheem's solution is incorrect.

Guys, relax! I got this!

http://www.wolframalpha.com/input/?i=sqrt+x+%3D+x%5E2+-+4x+%2B+2

Unfortunately Raheem's solution is incorrect.

Can you explain how i am wrong?

- yours = same as britmillers = correct

Raheem's correct, because a function will only intersect with it's inverse on the line y = f(x).

Anyways, found all the solutions but they are all < 2 therefore making x = 4 the only valid solution. The only problem with the solutions < 2 is that you have to use the negative square root for them to be true when replaced into the function. So essentially, does the question only expect me to find one solution by trial and error which is simply 4?

How?

How am i wrong according to the graph?

The other solution isn't correct, you need to take the domain of x>2, otherwise the inverse doesn't exists.

Indeed, my apologies.

Having given Raheems method more thought, it is obviously correct.

The inverse of a function is the reflection of a function in the line y=x, as Raheem said. Therefore when the two are equal, they must both lie on y=x, so they are invariant to an inversion in the line y=x (as if they weren't then they wouldn't be equal).

Therefore you can set either the function, or indeed, it's inverse equal to x.

That is:

$f^{-1}(x) = x \Rightarrow \sqrt{x} = x-2 \Rightarrow x^2 -5x +4 =0 \Rightarrow (x-1)(x-4) = 0 \Rightarrow x_{1,2}=1,4$

or

$f(x) = x \Rightarrow (x-2)^2 = x \Rightarrow x^2 - 5x +4 = 0 \Rightarrow (x-1)(x-4) = 0 \Rightarrow x_{1,2}=1,4$

As $x \geq 2$ for f(x) then the only valid solution is x=4.

Indeed, my apologies.

Having given Raheems method more thought, it is obviously correct.

The inverse of a function is the reflection of a function in the line y=x, as Raheem said. Therefore when the two are equal, they must both lie on y=x, so they are invariant to an inversion in the line y=x (as if they weren't then they wouldn't be equal).

Therefore you can set either the function, or indeed, it's inverse equal to x.

That is:

$f^{-1}(x) = x \Rightarrow \sqrt{x} = x-2 \Rightarrow x^2 -5x +4 =0 \Rightarrow (x-1)(x-4) = 0 [br]\therefore x_{1,2}=1,4$

or

$f(x) = x \Rightarrow (x-2)^2 = x \Rightarrow x^2 - 5x +4 = 0 \Rightarrow (x-1)(x-4) = 0[br]\therefore x_{1,2}=1,4$

As $x \geq 2$ for f(x) then the only valid solution is x=4.

you are correct - for the given restrictions on the domain - not otherwise

Could you explain why it lies on the reflection of y = x, I don't really understand.

Indeed, my apologies.

Having given Raheems method more thought, it is obviously correct.

The inverse of a function is the reflection of a function in the line y=x, as Raheem said. Therefore when the two are equal, they must both lie on y=x, so they are invariant to an inversion in the line y=x (as if they weren't then they wouldn't be equal).

Therefore you can set either the function, or indeed, it's inverse equal to x.

That is:

$f^{-1}(x) = x \Rightarrow \sqrt{x} = x-2 \Rightarrow x^2 -5x +4 =0 \Rightarrow (x-1)(x-4) = 0 [br]\therefore x_{1,2}=1,4$

or

$f(x) = x \Rightarrow (x-2)^2 = x \Rightarrow x^2 - 5x +4 = 0 \Rightarrow (x-1)(x-4) = 0[br]\therefore x_{1,2}=1,4$

As $x \geq 2$ for f(x) then the only valid solution is x=4.

you are correct - for the given restrictions on the domain - not otherwise

Raheem's correct, because a function will only intersect with it's inverse on the line y = f(x).

That is not true, look at the graphs that hasufel attached.