solve the equation (3x+2)(x-4)=0 enter the solution with the highest value.

Using completing the square:
(3x+2)(x-4)=0
3x^2 - 12x + 2x - 8 = 0
3x^2 - 10x = 8
3(x^2 - 10x/3) = 8
3((x - 10/6)^2 - (10/6)^2) = 8
3((x - 5/3)^2 - (5/3)^2) = 8
3((x - 5/3)^2 - 25/9) = 8
3(x - 5/3)^2 - 25/3 = 8
3(x - 5/3)^2 = 49/3
(x - 5/3)^2 = 49/9
x - 5/3 = ±√(49/9)
x - 5/3 = ±(7/3)
x = (5/3) ± (7/3)
x = 12/3, -2/3
x = max(12/3, -2/3) = 12/3
x = 4

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cecylya cecylya · Answer 2 · 2019-08-19T00:52:08+02:00

Answer:

The correct answer is x=4

Step-by-step explanation:

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

3x ^2-12x+2x-8=0

3x ^2-10x-8=0

We can use Quadratic Formula.

If we have an equation with the following Standard Form , ax2 + bx + c = 0

(a, b and c are known values. a can't be 0.)

"x" is the unknown

we can solve the equation applying the Quadratic Formula.

x = (−b ± √(b^2 − 4ac))/ 2a

In this case,

a=3

b=-10

c=-8

Replacing

x = (10 ± √(-10^2 − 3*2*-8))/ 2*3

x=10 ± √(100 − (-96))/ 6

x=10 ± (√196)/6

x=(10 ± 14)/6

We have 2 answers

1- x=(10 + 14)/6=24/6=4 this is the highest value

2- x=(10 - 14)/6=-4/6=-0.66