Mathathon 2020 Problem 4 $-$ Medium

$9x^2+12x+4=x^3+3x^2(\sqrt[3]{25}-1)+3x(\sqrt[3]{25}-1)^2+(\sqrt[3]{25}-1)^3$ $(3x)^2+2(3x)(2)+2^2=x^3+3x^2(\sqrt[3]{25}-1)+3x(\sqrt[3]{25}-1)^2+(\sqrt[3]{25}-1)^3$ $(3x+2)^2=(x+\sqrt[3]{25}-1)^3\quad\blue{\because a^2+2ab+b^2=(a+b)^2 ;\space a^3+3a^b+3ab^2+b^3=(a+b)^3}$ $3x+2=(x+\sqrt[3]{25}-1)^\frac{3}{2}........[1]$ Let $\alpha = x+\sqrt[3]{25}-1$ $\Rightarrow x=\alpha - \sqrt[3]{25}+1..........[2]$ $\Rightarrow 3x+2=3\alpha-3\sqrt[3]{25}+3+2=3\alpha -3\sqrt[3]{25}+5$ Putting this in $[1]$ $3\alpha -3\sqrt[3]{25}+5=\alpha^\frac{3}{2}$ $3\alpha -3\sqrt[3]{25}+5=\alpha\sqrt{\alpha}$ $3\alpha-\alpha\sqrt{\alpha}=3\sqrt[3]{25}-5$

${\alpha}({3}-{\sqrt{\alpha}})={\sqrt[3]{25}}({3}-{\dfrac{5}{\sqrt[3]{25}}})$

$\boxed{\red{\alpha}({3}-\green{\sqrt{\alpha}})=\red{\sqrt[3]{25}}({3}-\green{\sqrt{\sqrt[3]{25}}})}\quad\blue{\because \frac{5}{\sqrt[3]{25}}=\frac{5}{5^\frac{2}{3}}=5^{1-\frac{2}{3}}=5^{\frac{1}{3}}=((5^{\frac{1}{3}})^2)^\frac{1}{2}={^+_-}\sqrt{5^{\frac{2}{3}}}=\sqrt{\sqrt[3]{5^2}}=\sqrt{\sqrt[3]{25}} }$ Comparing both the sides of the above equation we get $\alpha=\sqrt[3]{25}$ From $[2]$ $x=\red{\cancel{\sqrt[3]{25}}}-\red{\cancel{\sqrt[3]{25}}}+1=\boxed{1}$

The given equation can be as:

$\begin{aligned} 9x^2 + 12 x + 4 & = \left(x + \sqrt[3]{25} - 1\right)^3 \\ 9(x^2 - 2x + 1) + 30x - 5 & = \left((x - 1) + \sqrt[3]{25} \right)^3 \\ 9(x-1)^2 + 30(x-1) + 25 & = (x-1)^3 + 3 \sqrt[3]{25}(x-1)^2 + 3(\sqrt[3]{25})^2(x-1) + 25 & \small \blue{\text{Putting }x = 1} \\ 25 & = 25 \end{aligned}$

Therefore, $x = 1$ is a root and $|x| = \boxed 1$ .

$\text{The given equation can be divided into two functions.}$

$f(x) = 9x^2 + 12x +4$ $g(x) = x^3 + 3x^2 \cdot (\sqrt[3]{25} - 1) + 3x \cdot (\sqrt[3]{25} -1)^2 + (\sqrt[3]{25} - 1)^3$ $z = \sqrt[3]{25} - 1)^3$

$\text{Notice:}\quad 4 < \sqrt[3]{25} - 1)^3$

$f'(x) = 18x + 12$ $g'(x) = 3x^2 + 6xz+3z^2$

$\text{ As z>4 and one of our functions has an uneven degree, the other one an even, we can deduce that our graphs will have intersected at least twice until the x-value that is the solution of f'(x)<g'(x).}$

$f'(x) \le g'(x)$

$18x + 12 \le 3x^2 + 11,4 + 10.83$ $22,83 \le-18x + 3x^2 - 11.4x$ $0 \le 3x^2 - 11,4x - 22.83$ $x_1 \approx 1.44; x_2 \approx -5.24$

$\text{Of course the solution may be outside the interval [-infinity; 1.44], but it may be worth taking a shot as we have a chance of 0.33 that we find our solution in that interval.}$

Luckily we can make that interval even smaller as x^3 is known to have a lot of negative y-values and f(x) can not have negative values for y.

$\text{To be exact, we search for a value of x where g(x) is switching signs.}$

$0 = x^3 + 3x^2 \cdot (\sqrt[3]{25} - 1) +3x \cdot (\sqrt[3]{25} - 1)^2 + (\sqrt[3]{25} - 1)^3$ $x \approx -1.924$

$\text{Now that we know that atleast one solution must be inside the interval [-1.924; -1.44].}$ $\text{Knowing that, we can make a chart with all values of g(x) inside the interval.}$ $\text{ We will only compute integer values of x as the problem solution is stated to be an integer.}$

$\text{25 is the only integer solution g(x). As f(x) must always have integer solutions we can now check if the point A(1,25) is a point of f(x).}$

$f(x) = 9x^2 + 12x + 4$ $f(1) = 9 \cdot 1 +12 \cdot 1 + 4$ $f(1) = 25$

$\text{ Answer: x=1 is the integer solution to the equation.}$

$\text{\large{Logic way:}}$

If you try a decimal in the answer box, then you'll see it only accepts integers.

• $\text{ When we try } x = 0$ :

$0 + 0 + 4 = 0 + 0 + 0 + 24$

$\text{4 and 24 are apart by 20 units}$

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• $\text{When we try } x = 1$

$\text{we get 25 and 61}$

$\text{they are apart by 36 units.}$

$\text{This distance is more than before, which means we are going the wrong way.}$

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• $\text{We try } x = -1$

$\text{we get} -1 = -1$

$\huge{|-1| = 1}$

YAY!

After some rearrangement, the equation reduces to the form :

$x^3+3(\sqrt[3] {25}-4)x^2+3\left ((\sqrt[3] {25}-1)^2-4\right )x+\left ((\sqrt[3] {25}-1)^3-4\right )=0$ .

This is an equation of degree $3$ , and hence has $3$ roots . So something specific must be mentioned about the nature of the root wanted as the answer.

It seems from the answer box that an integer root is wanted.

That root of the equation is $\boxed 1$ .

So first we factor the RHS. Study the coefficients closely . This is the $3^{rd}$ row of the Pascal’s triangle . So $9x^2+12x+4=(x+(\sqrt[3]{25}-1))^3.$ Then the LHS. It has one zero(root): $\dfrac{2}{3}$ . The coefficient of $x^2$ is 9. So $9(x+\dfrac{2}{3})^2= (x+(\sqrt[3]{25}-1))^3\Rightarrow\dfrac{9(x+\dfrac{2}{3})^2}{(x+(\sqrt[3]{25}-1))^3}.$ We want to take the -1 from the $\sqrt[3]{25}-1$ away to make $25$ after cubing. So we test $x=1$ .
$\large \text{Bingo! It matches!}$
So the answer is $\boxed{1}$ .