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?!) Aug 27 '15 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. Aug 27 '15 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. Aug 27 '15 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.) Aug 29 '15 at 12:30
• I think it's best to ask a new question rather than edit a question that's already been answered. Sep 22 '15 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
Aug 27 '15 at 7:05
• Thanks for your comment-I guess it depends on personal preference. I do like Mark's method too. Aug 27 '15 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? Aug 27 '15 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. Aug 27 '15 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) Aug 27 '15 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.