Solving a 4th Degree Equation with Ease

[evagelos]

Homework Statement

Solve the following equation:

Homework Equations

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

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??

 

[ocohen]

have you tried factoring? look at the first equation

 

[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??

 

[ocohen]

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

 

[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

 

[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.

 

[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.

 

[evagelos]

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

thanks

 

[Redbelly98]

Homework Statement

Solve the following equation:

Homework Equations

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

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.

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

 

[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

 

[Mark44]

have you tried factoring? look at the first equation

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??

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.

 

[Fragment]

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

 

[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.

 

[Redbelly98]

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

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

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

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.

 

[Dickfore]

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.

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:

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

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

So, your equation breaks down to two quadratic equations:

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

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

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

 

[ehild]

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.

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

ehild