How to solve equations to the fourth power?

Is it possible to manually retrieve the value of $y$ from the following equation $$\color{blue}{153y^2-y^4=1296}$$

WolframAlpha has four solutions for $y$: $-12, -3, 3, 12$. How has it solved?

What I've achieved to until now is the following: $$y^2(153-y^2)=1296$$ And... I'm stuck.

• This is a REALLY easy quartic and not well researched. It invites you to consider $x=y^2$ then solve $-x^2+153x-1296$ which is easy. Solving quartics in general is not like this. (so who among you is upvoting and staring this?!) Commented Aug 27, 2015 at 8:53
• @AlecTeal OP showed attempt, the question is well formatted and if you are just getting used to the quadratic equations then this requires insight, but of course, when you know the trick its easy. Don't judge on how easy it is when you don't know the OPs level. Commented Aug 27, 2015 at 15:20
• @AlecTeal In my opinion that is unnecessarily rude. This site is for people of any mathematical ability who may not spot things as easily as you do. Commented Aug 27, 2015 at 18:20
• @imulsion Here is a more sophisticated question of similar quality with 1/3 the upvotes (currently). This one is nearly a duplicate with 1/5 the upvotes. The vagaries of SE voting are renowned. (I do agree, however, that the AlecTeal's tone appears rude, and I think the argument that the question should be downvoted, implied IMO by the assertion that it is "not well researched," is off-base for the reason you cite.) Commented Aug 29, 2015 at 12:30
• I think it's best to ask a new question rather than edit a question that's already been answered. Commented Sep 22, 2015 at 12:40

One method is to set $x=y^2$ and rearrange this as a quadratic equation $$x^2-153x+1296=0$$

Here's the solution:

$$y^4-153y^2 +1296 = 0$$

$$y^4 -144y^2-9y^2+1296 = 0$$

$$y^2(y^2-144) -9(y^2-144) = 0$$

$$(y^2-9)(y^2-144)=0$$ $$(y^2-3^2)(y^2-12^2)=0$$

note that $a^2 - b^2 = (a-b)(a+b)$

Can you see how now? I trust you can finish the rest

• Nice find. The method by Mark Bennet is much more robust, as it does not depend on seeing the factorisation, but this is much more beautiful :)
– 5xum
Commented Aug 27, 2015 at 7:05
• Thanks for your comment-I guess it depends on personal preference. I do like Mark's method too. Commented Aug 27, 2015 at 7:07
• @John - of course it works, but how did you quickly notice the middle term can be split like that? Or rather, how long did it take you to figure it? Commented Aug 27, 2015 at 13:03
• @JoeTaxpayer, the only way to figure that out is trial and error. 1296 factors to $2^4 \times 3^4$, 153 is an odd number so you need one odd and one even factor-hence one factor can only be a multiple of 3. Commented Aug 27, 2015 at 13:06
• @John - "look for 2 numbers that multiply to 1296, and add to -153" - that would be the advice for my students, right? Followed by factoring the 1296 to see the candidates. (thx) Commented Aug 27, 2015 at 13:10

While the convert-to-quadratic approach works here, and other answers have provided this approach, it's worth noting that rational solutions to quartics with integer coefficients work similarly to quadratics.

That is, if you have a quartic of the form $$ax^4+bx^3+cx^2+dx+e=0$$ where all coefficients are integers, then all rational solutions must be of the form $$x = \frac{\text{divisor of }e}{\text{divisor of }a}$$ In this case, we have $a=1$, $c=-153$, $e=1296$, and $b=d=0$. So we seek a solution of the form $$x = \frac{\text{divisor of }1296}{\text{divisor of }1}$$ As the only divisors of 1 are 1 and -1, we need only look at integer values. 1296 has the prime factorisation $$1296 = 2^4\times 3^4$$ and from here, we just need to start substituting trial values in until we find a root. With $P(x)=x^4-153x^2+1296$, noting that only even powers of $x$ appear (and thus we need not consider the negatives separately), we have $$P(1) = 1 - 153 + 1296 = 1144 \neq 0\\ P(2) = 16 - 612 + 1296 = 700 \neq 0\\ P(3) = 81 - 1377 + 1296 = 0$$ and thus $x=3$ is a solution. Again noting that only even powers appear, we see that $x=-3$ is also a solution. It's fairly trivial to continue in this way to find the other two roots, as well.

Note that this method won't find irrational roots, however.