Is Squaring the Best Method to Solve Absolute Value Quadratic Equations?

[Hootenanny]

I need to solve:
$$\left| 2x-3 \right| = 5 - x^2$$
I started by squaring the equation because modulus functions can only be positive and obtained:
$$x^4 -14x^2 +12x +1 = 0$$
I haven't olved any quadnomial equations before, so I don't know where to start. Any help would be much appreciated.

 

[VietDao29]

I need to solve:
$$\left| 2x-3 \right| = 5 - x^2$$
I started by squaring the equation because modulus functions can only be positive and obtained:
$$x^4 -14x^2 +12x +1 = 0$$
I haven't olved any quadnomial equations before, so I don't know where to start. Any help would be much appreciated.

Nah squaring both sides just turns your little problem into a monster...
It'll help if you note that:
|A| >= 0.
So if you are saying that |A| = B, then is it obvious that B should be non-negative, too?
Now if B is non-negative, and we have |A| = B, so that means:
A = B, if A >= 0
A = -B, if A < 0, right?
So all you have to do is to solve the system of equations:
$$\left\{ \begin{array}{l} 5 - x ^ 2 \geq 0 \\ \left[ \begin{array}{l} 2x - 3 = 5 - x ^ 2 \\ 2x - 3 = x ^ 2 - 5 \end{array} \right. \end{array} \right.$$
Can you get it? :)

 

[Hootenanny]

I get:
$$x\leq\sqrt{5}$$
Then for the first equation $x=2$ and $x=-4$.
And for the second; $2\pm 2\sqrt{3}$ but we must ignore the $2 + 2\sqrt{3}$ because it lies outside the inequality.
Ive got three answers what have I done wrong?

 

[shmoe]

The split to the cases 2x-3=5-x^2 and 2x-3=x^2-5 had additional restrictions on x that came from the absolute value sign, you haven't taken this into account yet.

 

[VietDao29]

I get:
$$x\leq\sqrt{5}$$

This is wrong: What if x = -7, -7 < sqrt(5), but 5 - (-7)² = 5 - 49 = -44 < 0!
What you should get is:
$$-\sqrt{5} \leq x \leq \sqrt{5}$$

Then for the first equation $x=2$ and $x=-4$.

This is correct. :)
However, -4 is not a valid solution, it's outside the range.

And for the second; $2\pm 2\sqrt{3}$ but we must ignore the $2 + 2\sqrt{3}$ because it lies outside the inequality.
Ive got three answers what have I done wrong?

Nope, you've solved the second equation incorrectly. You forget to divide it by 2a (i.e: 2).
Can you go from here? :)
------------
By the way, one can always check their answer by plugging the solution back to the equation. For example: x = 2 is one of the solution. So:
|2x - 3| = |2 . 2 - 3| = 1
5 - x² = 5 - 2² = 1.
And hurray! 1 = 1.
Can you get this? :)

 

[VietDao29]

The split to the cases 2x-3=5-x^2 and 2x-3=x^2-5 had additional restrictions on x that came from the absolute value sign, you haven't taken this into account yet.

Nah, we just need the "restriction" 5 - x² >= 0. If: 2x - 3 = 5 - x², then it's obvious that 2x - 3 >= 0.
And if: 2x - 3 = -(5 - x²) = x² - 5, then it's obvious that 2x - 3 <= 0. No?
So one is enough, I think.

 

[Hootenanny]

Ahh yes, I remember the critical values now. It's along time since I've practised solving inequalities.

So
$$x = \frac{2\pm 2\sqrt{3}}{2} \Rightarrow x = \pm\sqrt{3}$$
Both solutions lie with the inequality so the solutions are $x=-4, x=-\sqrt{3},x = \sqrt{3}, x=2$

Does that look ok?

 

[VietDao29]

Ahh yes, I remember the critical values now. It's along time since I've practised solving inequalities.

So
$$x = \frac{2\pm 2\sqrt{3}}{2} \Rightarrow x = \pm\sqrt{3}$$

Nah, this is again wrong... :tongue2:
It should read:
$$x = 1 \pm \sqrt{3}$$ :)

Both solutions lie with the inequality so the solutions are $x=-4, x=-\sqrt{3},x = \sqrt{3}, x=2$

Does that look ok?

No, that does not, you may want to re-check that, there are up to 2 solutions that do not satisfy the inequality 5 - x² >= 0.
You'll have only 2 valid solutions left.
Can you go from here? :)

 

[Hootenanny]

O dear, does the lack of sleep show?

So the only valid solutions are:
$$x=2$$
$$x=1-\sqrt{3}$$

Thank's very much for your help.

 

[VietDao29]

O dear, does the lack of sleep show?

Pretty much.

So the only valid solutions are:
$$x=2$$
$$x=1-\sqrt{3}$$

Yes, this is correct.
Congratulations, :)

Thank's very much for your help.

It's my pleasure.

 

[sidhantdas]

can i have solution of the Question if f(x) = 1-x/1+x,x>0 then f(f(x)) +f(f(1/x))