# Just squaring won't work

1. Aug 5, 2014

### Dinheiro

1. The problem statement, all variables and given/known data
Solve the equation
(x² + x -2)² + x² = 4
*Calculators aren't allowed

2. Relevant equations
Algebra

3. The attempt at a solution
(x² + x -2)² + x² + (x - 2) = 4 + x - 2
Now, let's say y = x² + x -2, then
y² + y - (2 + x) = 0
But it isn't solvable yet

2. Aug 5, 2014

### Staff: Mentor

Have you tried plugging some numbers in it like 0 or 1?

or even some small negative numbers like -1 or -2?

To start solving it, try expanding the first term (x^2 + x - 2)^2 and then add in the x^2 term

use the trick:
Code (Text):

x^2   +    x    - 2
*  x^2   +    x    - 2
-----------------------
-2x^2    - 2x   + 4
x^3 +...
x^4 + ...
-----------------------------
x^4 + ...

Last edited: Aug 5, 2014
3. Aug 5, 2014

### ehild

That method does not help. Now you have two unknowns instead of one.
Expand the square, bring the equation in form f(x)=0 and try to factorize . It is quite simple.

ehild

4. Aug 5, 2014

### Mentallic

Take -4 on each side, then $x^2-4=(x-2)(x+2)$ and then add $x^2(x+2)$ to each side. It should all fall into place after that.

Also keep in mind that as jedishrfu suggested, you can find some easy rational roots such as x=0 and x=-2, and when you do find such roots, you then know that

$$f(x)=(x^2+x-2)^2+x^2-4=x(x+2)(ax^2+bx+c)$$

For some constant a,b,c (you can quickly tell that a=1 for example) and then you can find b and c after comparing coefficients. This is also what polynomial division accomplishes.

5. Aug 6, 2014

### Staff: Mentor

To the OP: to make it look less random, it is an application of the integral root theorem. If you have a polynomial, where the leading coefficient is 1, if the polynomial has an integer root, it is guaranteed to divide the free coefficient.

If you try to attack the problem by brute force, just squaring and tidying the polynomial in question you will find there is no free coefficient, so 0 is one of the roots. After factoring it out, you will find the remaining 3rd degree polynomial has the free coefficient equal 4 - so possible integer roots are -4, -2, -1, 1, 2 and 4. Just try them all, and you will be left with a simple quadratic equation.

This is in turn weaker version of the http://en.wikipedia.org/wiki/Rational_root_theorem.

6. Aug 6, 2014

### Ray Vickson

If this is an olympiad question, you might assume the solution is easy, but tricky. Try assuming the two terms on the left are equal; that is, try $x^2 = 2, \: x^2 + x - 2 = x$.

7. Aug 6, 2014

Thanks!

8. Aug 6, 2014

### Staff: Mentor

x2+x-2=(x+2)(x-1)

So
$$(x^2+x-2)^2=(x+2)^2(x-1)^2$$

Try substituting that into your equation, and factoring out (x+2). The remaining factor will simplify greatly if you expand it.

Chet

9. Aug 6, 2014

### Dinheiro

Thanks, Chet, a good way to solve too

10. Aug 7, 2014

### ehild

You could have solved the problem without any tricks, just expanding the square:

$(x^2+x-2)^2+x^2-4=0$
$x^4+2x^3+x^2-4x^2-4x+4+x^2-4=0\rightarrow x^4+2x^3-2x^2-4x=0$
x can be factored out, one root is zero.

$x^3+2x^2-2x-4=x^2(x+2)-2(x+2)$... easy to factorize further.

ehild