Can it be transformed into quadratic?

Algebra Level 3

$16x^{4} + 8x^{3} - 12x^{2} - 4x +1 =0$ If $a,b,c$ and $d$ are the real roots of the above equation such that $a<b<c<d$ , then find the value of $\left\lfloor \dfrac{100ab}{cd} \right\rfloor$ .

The answer is 353.

3 solutions

Pil Pinas
Jul 5, 2016

Factor out the polynomial

(2x+1)(8x^3-6x+1)=0

One of the roots is -1/2, you get also get this root easily through other methods (e.g. substitution)

Let's get the roots for 8x^3-6x+1=0

Substitute x for sin a.

8 (sin a)^3-6(sin a) +1=0

1=-8 (sin a)^3+6(sin a)

1/2= -4(sin a)^3+3(sin a)

By sine triple angle identity, (you can prove it for yourself)

1/2= sin 3a

So 3a= 30 + 360k or 150+ 360k (in degrees), where k is an integer

a= 10+ 120k or 50+120k

a= 10, 130, 250, 50, 170, 290 (other angles are just repetitions as they go periodic around the unit circle)

Note that sin10= sin 170, sin 50= sin 130, sin 250= sin 290

So we can say that our other roots are sin 10, sin 50, and sin 250

Plugging the roots, we get

(100)(-1/2)(sin250)/(sin10)(sin50)

=353.2088886

It's not a great way to factorise the polynomial......

Anubhav Mahapatra - 3 years, 10 months ago

A question sudden comes into my mind, which I didn’t think about when leaarning trig substitution in calculus. How can you make sure that the roots are bounded by 1 so that you can use the trig function to substitute them?

Yinchen Wu - 3 years, 3 months ago

you can't do sin sub. first you have to prove that the root lies between [-1,1] . then you can sub sin

Himanshya Bhardwaj - 9 months, 3 weeks ago

You can use the substitution $x=a\sin(t)$ or $x=a\cos(t)$ with $a\in\mathbb{R}$ instead - this may be the simplest way to find all real roots of the depressed cubic $x^3-px-q=0$ if you do not want to apply the cubic formula directly (see my solution below).

If you're comfortable with complex numbers, the substitution $x=z + \frac{p}{3}z^{-1}$ yields all three solutions, real and complex alike.

Carsten Meyer - 1 month, 3 weeks ago

Aneesh Kundu
Nov 8, 2014

Make the substitution $x=\frac{t}{2}$ and then divide by $t^2$ and complete the square to get the a quadratic in terms of $t-\frac{1}{t}$ . Solve it and then revert back to x.

What're the coefficients of the quadratic in $t-\frac{1}{t}$ and how do you reach it; please elaborate, thanks!

Alex Delhumeau - 5 years, 12 months ago

I can't repeat your procedure either. Making the first substitution reduces the equation to

t^4 + t^3 - 3 t^2 - 2 t + 1 = 0

Dividing by t^2 (since t does not equal 0) yields

t^2 + t - 3 - 2/t + 1/t^2 = 0

In terms of (t - 1/t), the square is (t^2 - 2 + 1/t^2).

Collecting these terms leaves

(t - 1/t)^2 +t -1 -2/t

The remaining terms in t are unbalanced. Either we add and subtract t to get

(t - 1/t)^2 + 2 (t - 1/t) - t -1 = [(t - 1/t) + 1]^2 - t - 2 = 0

or we add and subtract 1/t to get

(t - 1/t)^2 + (t - 1/t) - 1/t - 1

Neither choice results in a clean quadratic in (t - 1/t), although the first one is close.Where do you go from here?

After the first substitution, I tried making it a polynomial in (t + 1), since the lead coefficient and the constant were both 1.

Since (t + 1)^4 = t^4 + 4 t^3 + 6 t^2 + 4 t + 1, I had to subtract 3 t(t^2 + 2 t +1) and 3 t(t + 1).

Then, (t + 1)^4 - 3 t(t + 1)^2 - 3 t(t + 1) = (t + 1) [(t + 1)^3 - 3 t(t + 1) - 3 t]

Clearly, one root for t is -1, so x = -1/2. It did not seem productive to continue in terms of (t + 1), so I expanded the terms in the cubic factor to get [t^3 - 3 t + 1]. This turned out to be unfactorable, but being in reduced cubic form, it was relatively simple to use the cubic formula to get three real roots for x (in degrees): cos 40, cos 160, cos 280. From there, the answer is 353

Tom Capizzi - 4 years, 12 months ago

Carsten Meyer
Apr 18, 2021

To compute the answer, we actually need to find the roots of the given polynomial $P(x)$ . Simplify it with the substitution $y:=2x$ : $0=P(x)=P(y/2)=y^4+y^3-3y^2-2y+1=:Q(y)\qquad|\text{Guess integer root }y=\red{-1}$ Divide $Q(y)$ by $(y+1)$ using Horner's Method: $\begin{aligned} \begin{array}{r|rrrrr} Q(y)&1 & 1 & -3 & -2 & 1\\\hline & \red{-1} & -1 & 0 & 3 & -1\\\hline R(y) & 1 & 0 & -3 & 1 & \underline{\underline{0}} \end{array} && \Rightarrow && Q(y) = (y+1)(y^3-3y+1) =:(y+1)R(y) \end{aligned}$ We note $R(\pm 1)\neq 0$ , so by Rational Root Theorem $R(y)$ cannot have rational roots we could guess: We need the cubic formula (or Cardano's Method ) to find the roots of $R(y)$ !

Luckily, $R(y)$ is already in depressed form $R(y)=y^3-py-q$ with $p=3,\:q=-1$ , so we can directly apply the cubic formula: $\begin{aligned} D&:=\left( \frac{p}{3} \right)^3-\left( \frac{q}{2} \right)^2=1-\frac{1}{4}=\frac{3}{4}>0 \qquad \Rightarrow\qquad \text{3 real solutions }y_{0,\:1,\:2}\\[1em] y_k&=2\sqrt{\frac{p}{3}}\cos\left( \varphi_0+\frac{2\pi k}{3} \right), \qquad\varphi_0=\frac{1}{3}\arctan2\left( \sqrt{D},\:\frac{q}{2} \right)=\frac{1}{3}\arctan2 \left( \frac{\sqrt{3}}{2},\:-\frac{1}{2} \right)=\frac{1}{3}\cdot\frac{2\pi}3{}=\frac{2\pi}{9} \end{aligned}$ With the substitution $y=2x$ , we can finally compute the four roots $x_k$ of $P(x)$ and order them into $a,\:b,\:c,\:d$ : $\begin{aligned} a &=\cos\left(\frac{8\pi}{9}\right), & b&=-\frac{1}{2},&c&=\cos\left(\frac{14\pi}{9}\right), &d&=\cos\left(\frac{2\pi}{9}\right) &&\Rightarrow &&& \left\lfloor\frac{100 ab}{cd}\right\rfloor = \boxed{353} \end{aligned}$

Rem.: As all roots are of the form $\cos(2\pi k/9)$ , there might be a more elegant solution than the cubic formula...

Can it be transformed into quadratic?

The answer is 353.

3 solutions

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