If α and β are zeroes of 2x^2-5x+7 find a polynomial whose zeroes are 2α +3β, 3α+2β

Question

It does not solve through middle tern split pls help

Answer 1

2x² - 5x + 7 = 0

Since this is not factorable use quadratic equation.

x = (-b ±√(b²-4ac))/2a  with a=2, b=-5, c=7

 

x = [5±√((-5)²-4(2)(7))]/(2(2))

 

x = [5 ±√(25-56)]/4

 

x = (5 ± i√31)/4

 

a = (5+i√31)/4

b = (5-i√31)/4

 

2a = (5+i√31)/2

3b = (15 - 3i√31)/4

 

2a + 3b = (10 + 2i√31 + 15 - 3i√31)/4

            = (25 -i√31)/4

            = 25/4 - i√31/4

 

If 2a+3b is a zero then x - (2a+3b) is a factor

[x - (25 -i√31)/4] is a factor of p(x)

or

(x - 25/4 + i√31/4)

 

3a = (15+3i√31)/4

2b = (5-i√31)/2

 

3a + 2b = [15+3i√31+10-2i√31]/4

            = (25 + i√31)/4

            = 25/4 + (i√31)/4

 

Then:  x - (3a+2b) =

          (x - 25/4 - (i√31)/4) is a factor of p(x)

 

p(x) = (x - 25/4 + i√31/4)(x - 25/4 - (i√31)/4)

 

As you can see this is not going to multiply out nicely:

 

p(x) = x² - (25/2)x + 189/4 - (25i√31)/8

 

Are you sure there isn't a mistake in the problem statement?