+0

-1
403
2

Find the roots of this polynomials.

9x4- 12x3+ 28x2 - 16x + 16

Jan 25, 2018

#1
+1

Solve for x:

9 x^4 - 12 x^3 + 28 x^2 - 16 x + 16 = 0

Write the left hand side as a square:

(3 x^2 - 2 x + 4)^2 = 0

Take the square root of both sides:

3 x^2 - 2 x + 4 = 0

Divide both sides by 3:

x^2 - (2 x)/3 + 4/3 = 0

Subtract 4/3 from both sides:

x^2 - (2 x)/3 = -4/3

x^2 - (2 x)/3 + 1/9 = -11/9

Write the left hand side as a square:

(x - 1/3)^2 = -11/9

Take the square root of both sides:

x - 1/3 = (i sqrt(11))/3 or x - 1/3 = 1/3 (-i) sqrt(11)

x = 1/3 + (i sqrt(11))/3 or x - 1/3 = 1/3 (-i) sqrt(11)

x = 1/3 + (i sqrt(11))/3         or          x = 1/3 - (i sqrt(11))/3

Jan 25, 2018
#2
+1

Guest, this has a degree of 4 so there are 4 roots, some may be double or multiple roots of course.

The graph is concave up because the leading coefficent is +9 (positive)

$$9x^4-12x^3+28x^2-16x+16$$

I do not think guest has explained about writing this as a square.

I suspect he used a calculator program to do that.

However I can see that 9x^4 and 16 are both perfect squares so it is not unreasonable to examine the squared possibility.

It would have to be of the form    $$(3x^2+ax+4)^2$$   if it is a perfect square.

$$(3x^2+ax+4)^2\\ =3x^2(3x^2+ax+4)+ax(3x^2+ax+4)+4(3x^2+ax+4)\\ =9x^4+3ax^3+12x^2+3ax^3+a^2x^2+4ax+12x^2+4ax+16\\ =9x^4+3ax^3+3ax^3+12x^2+a^2x^2+12x^2+4ax+4ax+16\\ =9x^4\qquad+6ax^3\qquad+(24+a^2)x^2\qquad\qquad+8ax\qquad+16\\$$

6a=-12 so a =-2

so

24+(-2)^2=28  good

8a=-16            good

so

$$9x^4-12x^3+28x^2-16x+16$$ = $$(3x^2-2x+4)^2$$

So the roots would be the positive and negative of the answers that our guest found.

$$x =\pm\left(\frac{ 1+ i \sqrt{11}}{3}\right) \qquad or \qquad x =\pm\left(\frac{ 1- i \sqrt{11}}{3} \right)$$