# Solving an equation

1. May 31, 2010

### evagelos

1. The problem statement, all variables and given/known data

Solve the following equation:

2. Relevant equations

$$(x^2+6x+9)^2 +(x^2+4x+6)^2 = (x^2+8x+12)^2$$

3. The attempt at a solution

if we expand we will end up with an equation of 4th 3rd 2nd degree equation ,which will, if not impossible very difficult to solve.

Any ideas for easier way??

Last edited: May 31, 2010
2. May 31, 2010

### ocohen

have you tried factoring? look at the first equation

3. May 31, 2010

### evagelos

The 1st equation is a perfect square :(x+3)^2 ,the 2nd has a complex root ,the 3rd is :(x-2)(x-6) does this help??

4. May 31, 2010

### ocohen

sorry no. I'm not sure. It has nice roots though so that suggests there should be a nice solution

5. May 31, 2010

### ehild

Rearrange:

$$(x^2+4x+6)^2- (x^2+8x+12)^2= -(x^2+6x+9)^2$$

and factorize the left side of the equation.

ehild

6. May 31, 2010

### DJsTeLF

Not sure if this will help but maybe you could try expanding it all, gathering terms, cancel what you can and then attempt to re-factorise. Worth a shot I'd say.

7. May 31, 2010

### n.karthick

I will consider like this

$$(x^2 + ax+b)^2=x^4+2ax^3+(a^2+2b)x^2+2abx+b^2$$

Now you have three quadratic parts. Instead of expanding and simplifying each one of them you can directly write the final expression as

$$x^4+2(a_1-a_2+a_3)x^3+(a_1^2+2b_1-a_2^2-b_2+a_3^2+2b_3)x^2+2(a_1b_1-a_2b_2+a_3b_3)x+b_1^2 -b_2^2 +b_3^2 =0$$

You can substitute the values of $a_1, b_1$ etc from the given expression and solve the resulting equation.

8. May 31, 2010

thanks

9. Jun 2, 2010

### Redbelly98

Staff Emeritus
Actually, it is very easy to solve using the Rational Roots Theorem, as all solutions turn out to be rational.

10. Jun 2, 2010

### ehild

I hope evagelos solved this problem long ago, so I continue my post.
The equation is rearranged

$$(x^2+4x+6)^2- (x^2+8x+12)^2= -(x^2+6x+9)^2$$

and the left side factorized

$$(x^2+4x+6+x^2+8x+12)(x^2+4x+6-x^2-8x-12)= -(x^2+6x+9)^2$$

and simplified

$$(2x^2+12x+18)(-4x-6)= -(x^2+6x+9)^2 \rightarrow -4(x^2+6x+9)(2x+3)= -(x^2+6x+9)^2$$

One possibility is that

$$(x^2+6x+9)^2=0 \rightarrow x=-3$$

The other one:

$$4(2x+3)= x^2+6x+9 \rightarrow x^2-2x-3=0$$

with roots -1 and 3.

ehild

11. Jun 2, 2010

### Staff: Mentor

There is only one equation, so it doesn't make sense to talk about the "first" equation. You are really talking about the first expression on the left side of the equation.

12. Jun 2, 2010

### Fragment

Unless I am wrong, the equation has another solution, x=-3.

13. Jun 2, 2010

### Dickfore

The coefficients of all the quadratic trinomials are in a ratio 2 : 3. So, if you introduce the substitution:

$$y = 2x + 3$$

the equation turns to:

$$(x^{2} + 3 y)^{2} + (x^{2} + 2 y)^{2} = (x^{2} + 4y)^{2}$$

Squaring these is easier, since they are binomials. The result is:

$$x^{4} + 6 x^{2} y + y^{2} + x^{4} + 4 x^{2} y + y^{2} = x^{4} + 8 x^{2} y + y^{2}$$

which simplifies to:

$$y^{2} + 2 x^{2} y + x^{4} = 0$$

which, of course is a full square:

$$(y + x^{2})^{2} = 0$$

But, the square of a number is zero only if the number itself is zero (double root), so this is equivalent to:

$$x^{2} + y = 0$$

Substitute back the expression for y and you get a simple quadratic equation for x:

$$x^{2} + 2 x + 3 = 0$$

The discriminant of this equation is:

$$1^{2} - 1 \cdot 3 = 1 - 3 = -2 < 0$$

so there are no real solutions to the equation.

14. Jun 2, 2010

### Redbelly98

Staff Emeritus
Sorry, there is an error in this step. There should be terms 9y2, 4y2, and 16y2.

At the risk of repeating myself ...

Expand the expressions in the original equation, and collect terms to get a 4th degree polynomial. Then use the Rational Roots theorem to find potential rational solutions. I found there are 3 distinct ones, with one of them a repeated root. I have verified my three rational (actually integer) distinct roots in the original equation.

This is basic high school algebra. While the proposed solution tricks may be slick or elegant, the "brute force" method of factoring a polynomial is rather simple in this particular problem.

15. Jun 2, 2010

### Dickfore

Lol. Of course. So, what does it become after this correction:

$$^{4} + 6 x^{2} y + 9 y^{2} + x^{4} + 4 x^{2} y + 4 y^{2} = x^{4} + 8 x^{2} y + 16 y^{2}$$

$$3 y^{2} - 2 x^{2} y - x^{4} = 0$$

This discriminant of this quadratic equation (with respect to y) is:

$$(-x^{2})^{2} - 3 \cdot (-x^{4}) = x^{4} + 3 x^{4} = 4 x^{4}$$

is a complete square, so the equation has rational roots:

$$\frac{x^{2} \pm 2 \, x^{2}}{3} = \left\{ \begin{array}{l} x^{2} \\ -\frac{x^{3}}{3} \end{array}\right.$$

and can be factorized as:

$$3 (y - x^{2}) \, (y + \frac{x^{2}}{3}) = 0$$

$$(x^{2} - y) \, (x^{2} + 3 y) = 0$$

$$x^{2} - 2 x - 3 =0$$

$$x^{2} + 6 x + 9 = 0$$

These have integer roots (the second one is a perfect square and has a double root).

16. Jun 3, 2010

### ehild

I like this "brute force", thank you Redbelly!

ehild