I saw this problem:If $x^2-16\sqrt x =12$ what is the value of $x-2\sqrt x$?

To make notations simpler Let $x=t^2$ and $t^4-16t=12$ and $l= t^2-2t$.

I tried to use the fact the $\frac{12} l =\frac{t^4-16t}{t^2-2t}= t^2+2t +4 + \frac{8}{t-2}= (t-2)^2 + 6t+ \frac{t}{8l}= \frac{l^2}{t^2}+6\frac{l}{t-2}+ \frac{t}{8l}$

I tried to find some similarities between the $l$ cubic polynomial and the original quartic polynomial but I didn't find anything useful, needless to say that solving the quartic polynomial is not an option as the general solution is too complicated. Btw the answer is $2$ and since this is a "clean" and "nice" answer for a quartic polynomial there must be some trick


3 Answers 3


Question Summary (to Make Easier to Reference in the Answer)

$$ \text{If }x^2-16\sqrt{x}=12 \text{ what is the value of }f=x-2\sqrt{x} \tag{Eq. 1}$$

Checking for Consistency by Plugging in the $f=x-2\sqrt{x}=2$

Check $ x - 2\sqrt{x}-2=0$ implies $(\sqrt{x}-1)^2=2+1=3$ so $\sqrt{x}=1+\sqrt{3}$. (Only the positive solution holds since $\sqrt{x} \ge 0$).

So then $x=(1+\sqrt{3})^2=1+2\sqrt{3}+3=4+2\sqrt{3}$.

Then $x^2=16+12+16\sqrt{3}=28+16\sqrt{3}$ and $x^2 - 16\sqrt{x}-12=(28+16\sqrt{3})-16*(1+\sqrt{3})-12=(28-16-12)+(16-16)\sqrt{3}=0$.

Thus indeed it must be so that $x-2*\sqrt{x}=2$.

Solving for $f=x-2\sqrt{x}$ starting with $-2\sqrt{x}$ and then simplifying further

To start, one can solve for $-2\sqrt{x}$ as follows (subtracting $x^2$ from the left and right of Equation 1, and then dividing both sides by $8$): $$ \frac{(x^2-16\sqrt{x})-x^2}{8} =\frac{(12)-x^2}{8} \underset{implies}\implies \tag{Eq. 2} $$ $$ \underset{implies}\implies -2\sqrt{x}=\frac{12-x^2}{8} \underset{implies}\implies x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8}\\ \underset{implies}\implies x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8} \tag{Eqs.3}$$

Now further simplify Equations 3, by multiply each side by $8$ and then by subtracting from each side $8*x$ so then: $$ x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8} \underset{implies}\implies 8x - 16\sqrt{x}=8x+12+x^2 \\ \underset{implies}\implies - 16\sqrt{x}=12-x^2 \\ \underset{implies}\implies x^2 - 16\sqrt{x}-12=0 \tag{Eq. 4}$$ $$\underset{implies}\implies (\sqrt{x})^4 - 16\sqrt{x}-12=0 \tag{Eq. 5} $$ The above Equation 5 is a solvable fourth-degree equation according to the link, which makes its solution techniques a little bit more unfamiliar than a second order polynomial.

I am hoping to update the answer with the referenced technique, but also this link can be used directly to solve the problem even before I provide step-by-step instructions on the solution. The link has the author's request for algebraic steps to solve the problem.

Alternatively, there are on-line solvers also for this equation as follows for instance:

solution for sqrt x

Since $\sqrt{x}>0$ and it is a irrational solution greater than zero, the only applicable solution from that is $\sqrt{x}=1+\sqrt{3}$ as was to be proven.

Checking Line-by-Line Using $\sqrt{x}=1+\sqrt{3}$, $x=2*(2+\sqrt{3})$, and $x^2=28+16*\sqrt{3}$

From Equation 3, with $x=(\sqrt{x})^2 =\left(1+\sqrt{3}\right)^2=1+3+2\sqrt{3}=2*(2+\sqrt{3})$ and $x^2=28+16*\sqrt{3}$ and $\sqrt{x}=1+\sqrt{3}$:

$$ -2\sqrt{x}=\frac{12-x^2}{8} \underset{implies}\implies -2\left(1+\sqrt{3}\right) =\frac{12-\left(1+\sqrt{3}\right)^4}{8} \\ =\frac{12-\left(4+2*\sqrt{3}\right)^2}{8} \\ =\frac{12-\left(28+16*\sqrt{3}\right)}{8}\text{ is correct! } {\unicode{x2714}} \tag{Eqs. 3a}$$ $$ x-2\sqrt{x}=\left(4+2\sqrt{3}\right)-2*\left(1+\sqrt{3}\right) =2\text{ is correct! } {\unicode{x2714}} \tag{Eqs. 3b} $$ $$ x - 2\sqrt{x}=\frac{(8x)+12-x^2}{8} \\ \underset{implies}\implies 2= \frac{\displaystyle (8*2*(2+\sqrt{3}))+12-(28+16\sqrt{3})} {\displaystyle 8} \\ \underset{implies}\implies 2= \frac{\displaystyle ((32+16\sqrt{3}))+12-(28+16\sqrt{3})} {\displaystyle 8} \\ 2= \frac{\displaystyle (32+12-28)} {\displaystyle 8} = \frac{\displaystyle (16)} {\displaystyle 8} = 2\text{ is correct! } {\unicode{x2714}} \tag{Eqs. 3c}$$

Solving the Quartic 4th Degreee Polynomial $(\sqrt{x})^4 - 16\sqrt{x}-12=0$ from Equation 5

From Equation 5, the solutions for $\sqrt{x}$ can be written as solutions to the Quartic 4th Degree Polynomial $(\sqrt{x})^4 - 16\sqrt{x}-12=0$.

Since the coefficient for the leading term $(\sqrt{x})^4$ is 1, then generally it is possible to seek two second-degree polynomials such that:

$$ \left((\sqrt{x})^2+a*(\sqrt{x})+b \right)* \left((\sqrt{x})^2+c*(\sqrt{x})+d \right)=0 \tag{Eq. 6}$$

Expanding Equaiton 6, for each coefficient of $(\sqrt{x})^n$, the expanded term can be compared to the coefficients from Equation 6 and Equation 5. To start with, compare the expanded coefficient element for the expanded $(\sqrt{x})^3$ as $(\sqrt{x})^3*(a+c)=0$, so immediately Equation 6 can be simplified (with $c=-a$) in Equation 7 as: $$ \left((\sqrt{x})^2+a*(\sqrt{x})+b \right)* \left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0 \tag{Eq. 7}$$

  1. Observe that in Equation 7, that the roles of $b$ and $d$ are entirely similar. Either $d>b$ so that $d=b_+$, and $b=b_-$. Or $d<b$ so that $d=b_-$ and $b=b_+$. However, because of the Commutative Law of Algebra for Multiplication arbitrarily the symmetry can be broken, with the larger value $b_+$ being placed on the factor on the left, namely, $\left((\sqrt{x})^2+a*(\sqrt{x})+b \right)$. And $b_-$ can be placed on the factor on the right, namely $\left((\sqrt{x})^2-a*(\sqrt{x})+d \right)$. Hence: $$ \left((\sqrt{x})^2+a*(\sqrt{x})+b_+ \right)* \left((\sqrt{x})^2-a*(\sqrt{x})+b_- \right)=0 \tag{Eq. 8}$$

  2. Now consider the coefficient equations for $(\sqrt{x})^2$ as $(\sqrt{x})^2*(b_+ + b_- - a^2)=0$ so $(b_+ + b_- - a^2)=0$.

  3. Now consider the coefficient equations for $(\sqrt{x})^1$ as $(\sqrt{x})^1*(a(b_- - b_+))=-16*(\sqrt{x})^1$ so $(a*(b_- - b_+))=-16$ so $(a*(b_+ - b_-))=16$ implying that $a>0$ since $b_+>b_-$ implies that $b_+-b_->0$, also a positive value.

  4. Now consider the coefficient equations for $(\sqrt{x})^0$ as $(\sqrt{x})^0*(b_+*b_-)=-12*(\sqrt{x})^0$ so $b_+*b_-=-12$.

  5. There are now three Equations, namely $(b_+ + b_- - a^2)=0$, $(a*(b_+ - b_-))=16$, and $b_+*b_-=-12$. And there are three unknowns to be solved for, namely $a$, $b_+$, and $b_-$; So the rearrangement of the three equations is accomplished in Equation 9 as follows:

$$ a=\frac{\displaystyle 16}{\displaystyle b_+ - b_-} =\frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}} \underset{implies}\implies \\ \underset{implies}\implies a^2= \left( \frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}} \right)^2=b_+ + b_- = b_+ - \frac{12}{b_+} \tag{Eq. 9a}$$ $$ \underset{implies}\implies 16^2=256=\left(b_+ - \frac{12}{b_+} \right)* \left( b_+ + \frac{12}{b_+} \right)^2= \\ =(4)*(8*8)=(4)*(8)^2 \tag{Eq. 9b}$$ $$\tag{Eqs. 9}$$ Since the number $256$ has a factor of $(4)^1$ and also has factors of $(8)^2$, substitute $(4)^1=\left(b_+ - \frac{12}{b_+} \right)^1$ and also $8^2=\left( b_+ + \frac{12}{b_+} \right)^2$ and then check for consistency.

It might seem like a slight of hand to make these assignments. However, consider that $128=(2*2)*(2*2*2)*(2*2*2)$, and also the left term, namely $\left(b_+ - \frac{12}{b_+} \right)$ is less than the right term, namely $\left( b_+ + \frac{12}{b_+} \right)$ because $b_+>0$.

Consider integer decompositions.

For instance, if the squares right term $\left( b_+ + \frac{12}{b_+} \right)$ were 1, then that would leave $\left( b_+ - \frac{12}{b_+} \right)=\frac{256}{1}$ being greater than $\left( b_+ + \frac{12}{b_+} \right)$, which cannot be, since $b_+>0$.

It is the same situation if $\left( b_+ + \frac{12}{b_+} \right)=4$ because that would lead to $\left( b_+ - \frac{12}{b_+} \right)=\frac{256}{4^2}=16$, with $16>4$.

Because of the prime number decomposition that $256=2^8*4$ that this works, as $256=2^8=(2^2)*\left(2*2*2\right)^2$.

$$ \left(b_+ - \frac{12}{b_+} \right)=4 \underset{implies}\implies (b_+)^2-4*(b_+)=12 \\ \underset{implies}\implies \left((b_+)-2 \right)^2=12+4=16 \underset{implies}\implies b_+=4+2=6 \\ \tag{Eqs. 10a}$$ $$ \left(b_- - \frac{12}{b_-} \right)=4 \underset{implies}\implies (b_-)^2-4*(b_-)=12 \\ \underset{implies}\implies \left((b_-)-2 \right)^2=12+4=16 \underset{implies}\implies b_-=-4+2=-2 $$ $$\tag{Eqs. 10b}$$ $$ \left(b_+ + \frac{12}{b_+}\right)=8=6+\frac{12}{6}=6+2\text{ is correct!} \tag{Eqs. 10c}$$ $$ a=\frac{\displaystyle 16}{\displaystyle b_+ - b_- } =\frac{\displaystyle 16}{\displaystyle (6) - (-2) } =\frac{\displaystyle 16}{\displaystyle 8 } = 2 \tag{Eqs. 10d}$$ $$ \tag{Eqs. 10} $$

Note that $b_+*b_-=(6)*(-2)=-12$ as was to be expected, as the greater value for $b_+$ is selected because $b_+>b_-$, and also the lesser value of the polynomial equation for $b_-$.

Solving the Second Degree Polynomial Derived from the 4th Degree Quartic

Finally, the polynomial can be chosen that has a positive real root. Really all polynomial combinations need to be considered. For a 4th degree polynomial like this one, there are two real roots where one is greater than zero and one is less. And there are another two that have the factor of $\sqrt{-1}$ which is not a solution for $\sqrt{x}$ which needs a positive real solution.

If the terms for $-1*\sqrt{x}a$ and $b=b_-$ then the root of the second degree polynomial below must be real and positive as to be desired. So Equation 7 is repeated with those conditions now here:

$$ \left((\sqrt{x})^2+a*(\sqrt{x})+b \right)* \left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0 \\ \underset{implies}\implies \left((\sqrt{x})^2-a*(\sqrt{x})+d \right)=0 \tag{Eq. 7}$$ $$ \left((\sqrt{x})^2-2*(\sqrt{x})-2 \right)=0 \\ \underset{implies}\implies \left(\sqrt{x}-1\right)^2=2+1=3 \\ \underset{implies}\implies \sqrt{x}=1+\sqrt{3} \text{ as was to be proven!} \tag{Eqs. 11}$$ To highlight the solution now for $\sqrt{x}$ from the 4th degree polynomial quartic $(\sqrt{x})^4 - 16\sqrt{x}-12=0$, Equation 11 has the solution as: $$ \boxed{\sqrt{x}=1+\sqrt{3} \text{ as was to be proven!} }\tag{Eq. 12} $$

Verifying the Consistency of the 4th Degree Polynomial Quartic

Now there is also consistency. From the online Equation solver, there is also consistency that verifies this result:

online equation solver tractability

Since $a$ is positive, and $b_+>b_-$, the online equation solver has results for $a$, $b_+$, and $b_-$, namely $a$ =2, $b_+=6$, and $b_-=-2$.

The three Equations, namely $(b_+ + b_- - a^2)=0$, $(a*(b_+ - b_-))=16$, and $b_+*b_-=-12$, can be tested against the online results to find any algebraic errors, applying $a$ =2, $b_+=6$, and $b_-=-2$. There is no mistake is in Equations 9, and the values given by the online Equation Solver are correct, also yielding the desired consistent result:

  1. $(b_+ + b_- - a^2)=0$ implies $6-2-4=0$, correct!
  2. $(a*(b_+ - b_-))=16$ implies $2*(6-(-2))=2*8=16$, correct!
  3. $(b_+*b_-)=-12$ implies $6*(-2)=-12$, correct!

So, now Equations 9 are carefully checked with Equations 13 and are found to be correct: $$\text{Checking Eqs. 9a:}\\ a=\frac{16}{b_+ - b_-}=\frac{16}{6-(-2)}=2 \text{ is correct!} \tag{Eq. 13a}$$ $$ \left( \frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}} \right)^2 =b_+ + b_- = b_+ - \frac{12}{b_+} \tag{Eq. 13b}$$ $$ \underset{implies}\implies \left( \frac{\displaystyle 16}{\displaystyle b_+ + \frac{12}{b_+}} \right)^2=$$ $$= \left( \frac{\displaystyle 16}{6 + \frac{12}{6}} \right)^2 =4=6-\frac{12}{6}=6-2=4 \text{ is correct!} \tag{Eq. 13c}$$ $$\tag{Eqs. 13}$$

  • 2
    $\begingroup$ Is there a Reader's Digest version of this? $\endgroup$ Apr 2 at 10:29
  • 1
    $\begingroup$ @GerryMyerson: Equations 1 to 5 get to the forth-degree-polynomial in $\sqrt{x}$. Eq. 1 is a repeat of the question to make the answer self-contained (useful to print with stackprinter). Equation 5 can be symbolically completed by computer computation, and you are done with the Reader Digest's version. But that misses the point of the answer. The question author being scared of and then the answer author solving by hand the forth-degree polynomial is the heart of the problem. Also, carefully checking the answer is important to be convincing and avoid simple mistakes. $\endgroup$ Apr 9 at 10:47

$$\text{Let }y=\sqrt{x}$$$$y^4-16y-12=0$$$$(y^2-2y-2)(y^2+2y+6)=0$$$$y=1\pm\sqrt{3}\text{ or }y=-1\pm\sqrt{5}i\\$$$$\sqrt{x}=1\pm\sqrt{3}\text{ or }\sqrt{x}=-1\pm\sqrt{5}i$$$$x=4\pm2\sqrt{3}\text{ or }x=-4\mp2\sqrt{5}i\\$$$$x-2\sqrt{x}=2$$$$\text{or}$$$$x-2\sqrt{x}=-2\mp4\sqrt{5}i$$

  • $\begingroup$ Your approach of taking $y=\sqrt{x}$ and finding the corresponding polynomial is very similar to mine, except I just reference the term directlty $\sqrt{x}$ to various powers in the polynomial. Indeed we get at the same polynomial. I used symbolic tools to simplify the polynomial, and get its result. In any case, we arrive at the same polynomial, yours to start and mine with explanation, namely $(\sqrt{x})^4-16*\sqrt{x}-12=0$; and that polynomial is correct. The difficulty is at the end, you arrive at four possible answers for $\sqrt{x}$, whilst only one is a positive real number. $\endgroup$ Mar 24 at 7:47
  • $\begingroup$ I assume that in the third line, you need to replace $x$ by $y$, don't you? Can you explain how to get from the second to that third line? $\endgroup$
    – Dominique
    Mar 28 at 15:24

$$\begin{gathered} {x^2} - 16\sqrt x - 12 = 0 \hfill \\ * x \geqslant 0 \hfill \\ \Leftrightarrow {x^2} - 4x - 2x - 4\sqrt x - 12 + 6x - 12\sqrt x = 0 \hfill \\ \Leftrightarrow \left( {x + 2\sqrt x } \right)\left( {x - 2\sqrt x } \right) - 2\left( {x + 2\sqrt x } \right) + 6\left( {x - 2\sqrt x - 2} \right) = 0 \hfill \\ \Leftrightarrow \left( {x + 2\sqrt x } \right)\left( {x - 2\sqrt x - 2} \right) + 6\left( {x - 2\sqrt x - 2} \right) = 0 \hfill \\ \Leftrightarrow \left( {x - 2\sqrt x - 2} \right)\left( {x + 2\sqrt x + 6} \right) = 0 \Rightarrow x - 2\sqrt x = 2 \hfill \\ \end{gathered} $$


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