Question

slove 2x^2-3x^2-11x+6=0 given that the roots are in AP

Answer 1

2x3+3x2-11x-6=0

Three solutions were found :

x = 2

x = -3

x = -1/2 = -0.500

Step by step solution :

Step 1 :

Equation at the end of step 1 :

(((2 • (x3)) + 3x2) - 11x) - 6 = 0

Step 2 :

Equation at the end of step 2 :

((2x3 + 3x2) - 11x) - 6 = 0

Step 3 :

Checking for a perfect cube :

3.1 2x3+3x2-11x-6 is not a perfect cube

Trying to factor by pulling out :

3.2 Factoring: 2x3+3x2-11x-6

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -11x-6

Group 2: 2x3+3x2

Pull out from each group separately :

Group 1: (11x+6) • (-1)

Group 2: (2x+3) • (x2)

Answer 2

2x^2-3x^2-11x+6=0
4x-9x-11x+6=0
5x-11x+6=0
-6x+6=0
x=0