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Question: The equation $16x^4 + 16x^3 - 4x - 1 = 0$ has a multiple root. If $\alpha, \beta, \gamma, \delta$ ar...

The equation 16x4+16x3−4x−1=016x^4 + 16x^3 - 4x - 1 = 0 has a multiple root. If α,β,γ,δ\alpha, \beta, \gamma, \delta are the roots of this equation, then 1α4+1β4+1γ4+1δ4=\frac{1}{\alpha^4} + \frac{1}{\beta^4} + \frac{1}{\gamma^4} + \frac{1}{\delta^4} =

Answer

24

Explanation

Solution

We are given the quartic equation

16x4+16x3−4x−1=0.16x^4 + 16x^3 - 4x - 1 = 0.

Step 1. Finding a rational root:
Test x=12x=\frac{1}{2}:

16(12)4+16(12)3−4(12)−1=16⋅116+16⋅18−2−1=1+2−2−1=0.16\left(\frac{1}{2}\right)^4 + 16\left(\frac{1}{2}\right)^3 - 4\left(\frac{1}{2}\right) - 1 = 16\cdot\frac{1}{16}+16\cdot\frac{1}{8}-2-1 = 1+2-2-1=0.

Thus, x=12x=\frac{1}{2} is a root.

Step 2. Factorization:
Dividing the quartic by (x−12)(x-\frac{1}{2}) we get (after synthetic/division) the depressed cubic:

16x3+24x2+12x+2.16x^3+24x^2+12x+2.

Factor out 2:

2(8x3+12x2+6x+1).2(8x^3+12x^2+6x+1).

Now, test x=−12x=-\frac{1}{2} in the cubic:

8(−12)3+12(−12)2+6(−12)+1=−1+3−3+1=0.8\left(-\frac{1}{2}\right)^3+12\left(-\frac{1}{2}\right)^2+6\left(-\frac{1}{2}\right)+1 = -1+3-3+1=0.

So, x=−12x=-\frac{1}{2} is a root, and the cubic factors as (2x+1)(4x2+4x+2)(2x+1)(4x^2+4x+2). Notice that

4x2+4x+2=2(2x2+2x+1).4x^2+4x+2 = 2(2x^2+2x+1).

Thus the complete factorization is

16x4+16x3−4x−1=4(x−12)(2x+1)(2x2+2x+1).16x^4+16x^3-4x-1 = 4 \left(x - \frac{1}{2}\right)\left(2x+1\right)\left(2x^2+2x+1\right).

The roots are:

x=12,x=−12,x=−1±i2.x = \frac{1}{2},\quad x = -\frac{1}{2},\quad x = \frac{-1 \pm i}{2}.

Step 3. Computing 1α4+1β4+1γ4+1δ4\displaystyle \frac{1}{\alpha^4}+\frac{1}{\beta^4}+\frac{1}{\gamma^4}+\frac{1}{\delta^4}:
Let the roots be α=12\alpha=\frac{1}{2}, β=−12\beta=-\frac{1}{2}, γ=−1+i2\gamma=\frac{-1+i}{2}, δ=−1−i2\delta=\frac{-1-i}{2}.

  • For α=12\alpha=\frac{1}{2}:
1α4=1(12)4=16.\frac{1}{\alpha^4}=\frac{1}{\left(\frac{1}{2}\right)^4} = 16.
  • For β=−12\beta=-\frac{1}{2} (note the fourth power makes the sign positive):
1β4=16.\frac{1}{\beta^4}=16.
  • For γ=−1+i2\gamma=\frac{-1+i}{2}:
    Write in polar form: −1+i=2 ei3π/4-1+i = \sqrt{2}\,e^{i3\pi/4}. Thus,
γ=22 ei3π/4.\gamma=\frac{\sqrt{2}}{2}\,e^{i3\pi/4}.

Then,

γ4=(22)4ei3π=416⋅(−1)=−14,\gamma^4 = \left(\frac{\sqrt{2}}{2}\right)^4 e^{i3\pi} = \frac{4}{16}\cdot(-1) = -\frac{1}{4},

so

1γ4=−4.\frac{1}{\gamma^4} = -4.
  • Similarly, for δ=−1−i2\delta=\frac{-1-i}{2}, we also get:
1δ4=−4.\frac{1}{\delta^4} = -4.

Thus, the sum is

16+16+(−4)+(−4)=24.16+16+(-4)+(-4)=24.