asked 126k views
1 vote
I cant solve this question pls help

I cant solve this question pls help-example-1
asked
User Phresus
by
8.2k points

1 Answer

3 votes

Answer:

  • −1.40122230483
  • 0.770588502368
  • 4.63063380246

Explanation:

You want to find all three solutions to the equation x³ -4x² -4x +5 = 0.

Characteristics of the roots

The 2 changes in signs of the coefficients from left to right tells you there are 2 or 0 positive real roots. (Descartes' rule of signs.) Negating odd-degree terms and checking again tells you there is one negative real root.

The y-intercept is +5, and the left-side value at x=1 is -2. This tells you all of the roots are real, and one of them is between x=0 and x=1.

If the roots were rational, they would be ±1 or ±5, factors of the constant. None of these is a root, so we know all of the roots are irrational.

The good news is that we can find them a couple of ways:

  1. using any of several cubic formulas
  2. using iterative methods.

Newton's method

The formula for the "next guess" (x') in the iterative process using Newton's method is ...

x' = x -f(x)/f'(x)

where x is a root of f(x) = 0, and f'(x) is the first derivative of f(x).

We like to do this iteration using an interactive graphing calculator, because it can show us the next value of x even as we are typing the current value of x. In the attachment, the function f₁(x) is the iteration function. When its value is the same as its argument, we have found a solution. A good starting value is the x-intercept shown on the graph.

The solutions using Newton's method iteration are shown at the top of this answer, and in the attachment.

Trig solution

There is a relatively simple way to solve a cubic using trig functions that goes like this:

f(x) = x³ +ax² +bx +c = 0

p = -a/3

q = b -a²/3

r = a(2a² -9b)/27 +c

d = √(-4q/3)

h = 4r/d³

roots are ...

x = p + d·sin(arcsin(h)/3 +{-2π/3, 0, 2π/3}) . . . . . angles in radians

This will give the same result as above, in terms of trig functions of rational numbers. The trig functions themselves will be irrational.

(This is not a formulation you are likely to see in school. Something like it may show up in a math handbook. It really only works if there are 2 or 3 real roots, otherwise the trig functions have complex values.)

<95141404393>

I cant solve this question pls help-example-1
answered
User Ayelegun Kayode
by
7.3k points

No related questions found

Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.