Question

If alpha and beta are the roots of x2-2x+4=0 then alpha6-beta6 =

Answer 1

Please see the attachment

Answer 2

hi mate,

Solving x^2-2x+4 = 0 by Completing The Square .

Subtract 4 from both side of the equation :

x^2-2x = -4

Now the clever bit: Take the coefficient of x , which is 2 , divide by two, giving 1 , and finally square it giving 1

Add 1 to both sides of the equation :

On the right hand side we have :

-4 + 1 or, (-4/1)+(1/1)

The common denominator of the two fractions is 1 Adding (-4/1)+(1/1) gives -3/1

So adding to both sides we finally get :

x^2-2x+1 = -3

Adding 1 has completed the left hand side into a perfect square :

x^2-2x+1 =

(x-1) • (x-1) =

(x-1)^2

Things which are equal to the same thing are also equal to one another. Since

x^2-2x+1 = -3 and

x^2-2x+1 = (x-1)^2

then, according to the law of transitivity,

(x-1)^2 = -3

We'll refer to this Equation as Eq. #2.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-1^)2 is

(x-1)^2/2 =

(x-1)^1 =

x-1

Now, applying the Square Root Principle to Eq. #2.2.1 we get:

x-1 = √ -3

Add 1 to both sides to obtain:

x = 1 + √ -3

In Math, i is called the imaginary unit. It satisfies i^2 =-1. Both i and -i are the square roots of -1

Since a square root has two values, one positive and the other negative

x^2 - 2x + 4 = 0

has two solutions:

x = 1 + √ 3 • i

or

x = 1 - √ 3 • i

or......

Solving x^2-2x+4 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax^2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B^2-4AC

x = ————————

2A

In our case, A = 1

B = -2

C = 4

Accordingly, B^2 - 4AC =

4 - 16 =

-12

Applying the quadratic formula :

2 ± √ -12

x = —————

2

Both i and -i are the square roots of minus 1

Accordingly,√ -12 =

√ 12 • (-1) =

√ 12 • √ -1 =

± √ 12 • i

Can √ 12 be simplified ?

Yes! The prime factorization of 12 is

2•2•3

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 12 = √ 2•2•3 =

± 2 • √ 3

√ 3 , rounded to 4 decimal digits, is 1.7321

So now we are looking at:

x = ( 2 ± 2 • 1.732 i ) / 2

Two imaginary solutions :

x =(2+√-12)/2=1+i√ 3 = 1.0000+1.7321i

or:

x =(2-√-12)/2=1-i√ 3 = 1.0000-1.7321i

Two solutions were found :

x =(2-√-12)/2=1-i√ 3 = 1.0000-1.7321i

x =(2+√-12)/2=1+i√ 3 = 1.0000+1.7321i

i hope it helps you.