3x^2+4x+4=0 solve for x quadratic formula?

Answer 1

quadratic formula is (-b +- sqrt(b^2 - 4ac))/2a
a = 3
b = 4
c = 4

x = (-4 +- sqrt(4^2 - 4(3)(4)) / 2(3)
x = (-4 +- sqrt(16 - 48)) / 6
x = (-4 +- sqrt(-32)) / 6
x = (-4 +- i sqrt(32)) / 6
x = (-4 +- i sqrt(16)sqrt(2)) / 6
x = (-4 +- i4sqrt(2)) / 6
x = (4/6)(-1 +- i sqrt(2))
x = (2/3)(-1 +- i sqrt(2))

Answer 2

3x^2 + 4x + 4 = 0

The quadratic formula, for an equation ax^2 + bx + c = 0 is the following:
x = [ -b +/- sqrt(b^2 - 4ac) ] / (2a)

In this quadratic, a = 3, b = 4, and c = 4.

x = [ -4 +/- sqrt(16 - 4(3)(4)) ] / [ 2(3) ]
x = [ -4 +/- sqrt(16 - 48) ] / [ 6 ]
x = [ -4 +/- sqrt(-32) ] / 6

We're taking the square root of a negative number, which means we won't have real roots; we will have complex roots (which uses i, or sqrt(-1).

x = [ -4 +/- i sqrt(32) ] / 6
x = [ -4 +/- i sqrt(16*2) ] / 6
x = [ -4 +/- i sqrt(16)sqrt(2) ] / 6
x = [ -4 +/- i (4)sqrt(2) ] / 6
x = (-4 +/- 4i sqrt(2) ) / 6
Which we can reduce by dividing 2 from each term.

x = ( -2 +/- 2i sqrt(2) ) / 3

And a clean answer would be

x = (2/3) (-1 +/- i sqrt(2) )

Answer 3

____________________________

This is the quadratic formula...

Your equation is like this:
ax² + bx + c = 0

Where a, b and c are NUMBERS.

Then the solution for x is when you substitute the numbers to the formula:
x = [-b ± √(b² - 4ac)]/2a
____________________________

So now, the equation is
3x² + 4x + 4 = 0

Now,
a = 3
b = 4
c = 4
right?

Then we can substitute them to the formula:
x = [-b ± √(b² - 4ac)]/2a

and we get
x = [-4 ± √(4² - 4·3·4)]/(2·3)

Then, we simplify
x = [-4 ± √(16 - 48)]/6

We simplify some more
x = [-4 ± √(-32)]/6

We simplify the radical √(-32)
√(-32) = √[16 · (-2)] = 4√(-2) = 4√2 i

so we get:
x = (-4 ± 4√2 i)/6

Then we divide the numerator and denominator by 2 to simplify... then we get the final answer:
x = (-2 ± 2√2 i)/3

Therefore, the two solutions are:
x = (-2 + 2√2 i)/3
AND
x = (-2 - 2√2 i)/3

^_^

Answer 4

It is an impossible problem. -4 + or - the square root of the quantity 16-4(3)(4) all over 2 (3) is how you would set it up. 4(3)(4) =48. the square root of 16-48 is an impossible problem, which nullifies the entire thing.

Answer 5

There are a few ways you can solve this equation, but I recommend this:

QUADRATIC FORMULA:
x = -b [+ or - the square root] of (b^2 - 4ac) [all over] 2a.

Plug in your values if your equation is written in the form ax^2 + bx + c.

Answer 6

x = [- 4 ± √(16 - 48)] / 6
x = [- 4± √(- 32)] / 6
x = [- 4 ± 4i√(2) ] / 6
x = (2/3).[ - 1 ± i √2 ]