Old Man doing Algebra Needs Help!?

Question

I haven't attempted high-school Algebra since 1984 and I will soon take the Praxis test.

Can someone please help with this problem?

- Solve for x

3x^2 + 4g + 1 = 0

(please show and explain all of the steps).

Thanks!

Sorry, I made a mistake. 4g should have been 4x.

so the problem is actually: 3x^2 + 4x + 1 = 0

Thanks for all of your help!

Answer 1

3x^2 + 4g + 1 = 0

Subtract 1 from both sides

3x^2 + 4g = - 1

Subtract 4g from both side, or add a negative 4g (sometimes it help understand)

3x^2 = -4g - 1

Divide by 3

x^2 = (-4g - 1)/3

Now get rid of the square by square rooting both sides

x = Square root (-4g - 1)/3

It's a process of elimination to get x by itself on one side of the equal sign. You have to do it on both sides to keep the equation balanced. If the equation started with = 25g or something you would have to deal with.
I graduated in 1980 and love this site. It keeps me up on the math I haven't used in years. Good luck with the test.

Answer 2

Is 4g meant to be 4x?

3x² + 4x + 1 = 0
(3x + 1)(x + 1) = 0
x = - 1/3 , x = - 1

Answer 3

g is constant, presumably.

3x^2 = -4g - 1.

We might now surmise that g is a negative constant, otherwise we will be straying into complex numbers.

Next step x^2 = -(1/3)*[4g +1]

Finally x = sqrt{-(1/3)*[4g +1]}

With the implication is that the quantity inside the curly brackets is a negative number.

Answer 4

Subtract the 4g and the 1 from both sides to give you
3x^2= -4g-1.
Divide both sides by 3 to give you.
x^2= (-4g - 1)/3
Now square root both sides to get.
x= sqrroot((-4g-1)/3)