# Square rooting to make quadratic function

If I have $x^4 - 34x^2 + 225 = 0$, is it not possible to to square root both sides of the equation so that I now have $x^2 - 34x + 15$? If this is true, then how would I go about solving the equation from there? I've tried completing the square, but I keep winding up with a large sum not equal to zero when I check my answer. I can't imagine quadratic formula or another method would yield a different result, unless I am going about it wrong.

My professor in their lecture video on this problem instead used the substitution method however, by setting $y = x^2$, and thus $y^2 = x^4$, then replacing all the $x$'s with $y$'s. This method seems to be simply pulling things out of the air though, and doesn't look valid to me. If someone could explain it for me, that would be appreciated as well.

• The square root of $x^4+34x^2+225$ is NOT $x^2+34x+15$. I don't know how can you imagine such a horrifing thing. – Crostul Jun 7 '15 at 20:50
• Worst freshman's dream application ever, perhaps? – Chappers Jun 7 '15 at 20:51
• The thing splits nicely as $(x^2-9)(x^2-25)$. – André Nicolas Jun 7 '15 at 20:59

It's not pulling things out of the air, it's a simple substitution. When we introduce the variable $y = x^2$, we convert $$x^4 - 34x^2 + 225 = 0$$ which, as a quartic equation, we may not know how to solve (although this one happens to be easy to factor), into: $$y^2 - 34y + 225 = 0$$ which, as a quadratic equation, is simpler, and we have an equation for it. Notice that we didn't actually change the structure of the equation - we simply substituted a new variable. For our convenience.

That way, you can plug the equation for $y$ into the quadratic formula and get $y = 25$ or $y = 9$. But we're not solving for $y$, we're solving for $x = \pm\sqrt{y}$, so $x = \pm 5,\pm 3$.

Hint. $$x^4-34x^2+225=x^4-30x^2+225-4x^2=(x^2-15)^2-(2x)^2$$

• Why not just $(x^2-25)(x^2-9)$? – Barry Jun 7 '15 at 20:54
• It's certainly an inside-out way to complete the square. – Thomas Andrews Jun 7 '15 at 21:38

No, because $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$. For example, $$3 = \sqrt{9} = \sqrt{1+8} \neq \sqrt{1}+\sqrt{8}=1+2\sqrt{2},$$ so it's definitely false.

What you should do is consider it as a quadratic equation in $x^2$, which you can relabel as $y$ for convenience.

If you do square root both sides of the equation, you get $\sqrt{x^4-34x^2+225}=0$, which is not much easier than the original.

However, if you set $u=x^2$, then square both sides of the equation, you get $u^2=x^4$. Simply substituting this in will give $u^2-34u+225=0$, which admits solutions $u=9,25$. Now because $u=x^2,x=\pm\sqrt u$. Therefore the solutions are $\{\pm\sqrt9,\pm\sqrt {25}\}=\{-5,-3,3,5\}$.

You can check that all these solutions make the original equation true.

The video is doing essentially what you are trying to do. Your approach of just replacing all the $x^2$s with $x$s is not correct because in general $x^2 \neq x$. The video defines a new variable $y$ by $y=x^2$ and performs the substitution, getting $y^2-34y+15=0$ That looks like what you got, but uses the new variable to show it is different from $x$. Now you solve this quadratic by your favorite method, say by completing the square: $(y-17)^2-274=0, y=17\pm \sqrt {274}, x=\pm \sqrt {17 \pm \sqrt {274}}$ Your now have four solutions for $x$, as you should because the original equation was fourth degree. Alpha confirms the result. The solutions you are getting are probably the ones I have for $y$, which will not satisfy the equation if used for $x$

You should consider the inequality $\sqrt{a+kb+c}\neq \sqrt{a}+k\sqrt{b}+\sqrt{c}$, holding for almost every $4$-tuple of positive real numbers (and a whole bunch of others not necessairly positive).

So, you can try to square root both sides (which can only be done if both are non-negative...), but even so you won't get the equation you say. You would get something not particularly useful.

The procedure used by the teacher is a substitution, not "taking a square root". It means observing that $$x^4+34x^2+225=0\Longleftrightarrow \begin{cases}x^2=y\\y^2+34y+255=0\end{cases}$$

• So it's not distributing the square root to all the items in the left side of the equation, it's instead square rooting them all at once? – MrHost56 Jun 7 '15 at 20:54
• And I understand how substitution method works, just not why you can just pull something out of thin air like that. Why is possible for me to just set a new variable as relative to initial one? – MrHost56 Jun 7 '15 at 20:55
• 1) You cannot distribute the square root over a sum. 2) You are not square-rooting all at once: you are just solving the equation in $x^2$ (i.e. finding the possible values of $x^2$) and then retrieving what $x$ can be. – user228113 Jun 7 '15 at 20:58
• Okay so I get that then, but substitution method still eludes me. It's like I'm just making up something so the equation can work. – MrHost56 Jun 7 '15 at 21:13

you cannot just square root the coefficients and get the square root of a polynomial.

$(x^2 - 34x +15)^2 = x^4 - 34x^3 + 15x^2 -34x^3 + 576x^2 - 510x + 15x^2 - 576x + 255$

$= x^4 - 68x^3 + 606x^2 - 1086x + 255$ and we can see that the two polynomials are not equal.

If you have reason to believe that there is a repeated quadratic root to the polynomial you could say that $(ax^2 + bx + c)^2 = x^4 - 34x + 225$ and solve for a,b and c. But performing operations on individual coefficients almost always produces gibberish.

$(ax^2 + bx + c)^2 = x^4 - 34x + 225$

$a^2x^4 + 2abx^3 + (2acx^2 + b^2x^2) + 2 bcx + c^2 = x^4 - 34x + 225$

We can see that $a = 1$ and $c = 15$ almost immediately. we can also see that there is no cubic term on the right so $2ab$ must be zero. we know a must be 1 so $b = 0$ we assumed that it has a repeated quadratic square so we must check for a contradiction.

$(x^2 + 15)^2 = x^4 + 30x^2 = 225$ therefore there is no repeated quadratic root of the polynomial $x^4 - 34x + 225$