How to solve this ?

Question

Let a,b be the two solutions of the quadratic equation

 

      3x^{2 _{+ 6x + 7 =0}}

 

the value of (2a-b)(2b-a) is ?

Answer 1

If a and b are the solution of 3x² + 6x + 7 = 0, then

a + b = –6/3 = –2

    ab = 7/3

so that

(2a – b) (2b – a) = 5ab – 2a² – 2b²

                        = 5ab – 2(a² + b² + 2ab) + 4ab

                        = 9ab – 2(a +b)²

                        = 9(7/3) – 2(–2)²

                        = 13

This way, you don't need to know a and b explicitly, and still get the answer.

 

Now if you try to solve for a and b, you will find that a and b are complex numbers

a = –1 + (2/3)√3i

b = –1 – (2/3)√3i

so

(2a – b) (2b – a) = (–1 + 2√3i)(–1 – 2√3i)

                         = 1 + 12

                         = 13

 but this requires solving the given quadratic equation, working with complex numbers and more work. Not recommend!

            

Answer 2

[-6 +or- sqrt( 36 - 4(3)(7) ) ]/6

[-6 +or- sqrt( 36 - 84) ]/6

[-6 +or- sqrt( -48)]/6

[-6 +or- 4i*sqrt(3)]/6

-1 +or- (2/3)i*Sqrt(3)

-----------------------------------------------

A = -1 + (2/3)i*sqrt(3)

B = -1 - (2/3)i*sqrt(3)

2A = -2 + (4/3)i*sqrt(3)

2B = -2 - (4/3)i*sqrt(3)


2A - B = [-2 + (4/3)i*Sqrt(3) ] - [ -1 - (2/3)i*sqrt(3) ]
= -2 + (4/3)i*sqrt(3) + 1 + (2/3)i*sqrt(3)

= -1 + 2i*sqrt(3)


2b - a = -2 - (4/3)i*sqrt(3) - [ -1 + (2/3)i*sqrt(3) ]

= -2 - (4/3)i*Sqrt(3) + 1 - (2/3)i*sqrt(3)

= -1 - 2i*sqrt(3)

--------------------------------

 

(2a - b)(2a - b)=
1 - [ (-4)*3 ]=
1 - (-12)

13