Finding the Solution in a Quadratic Equation?

[Blue Leader]

Greetings,

I'm back with another few questions, unfortunately. There's a few problems that I'm stuck on, and I'm really not sure what to do with them...

I don't know if it makes any difference, but here are some of the instructions that were printed at the top of my paper...

Find the solution for each of the following equations.
1- Put the equation in \(\displaystyle Ax^2 + Bx + C = 0\) form
2- Determine the form of the equation
3- Find the solutions
-Set each factor equal to 0 and solve
-Solve using the square root method ( \(\displaystyle \sqrt{Ax^2 } = +- \sqrt{C}\))
-Solve by factoring
-Use the Quadratic Equation

Most of the problems are the \(\displaystyle Ax^2 + Bx + C = 0\) form, and I know that if there is nothing that adds to B and multiplies to C (or maybe it's the other way around) that I need to use the square root method, which is different for me as before we were just told to write "no solution". This time I guess we have to continue it.

I've got most of the problems solved, but there's still a few I'm struggling with and I'm not sure where to go with them. They're probably easier than I think they are, so I'm probably over-thinking things, but...

1. \(\displaystyle x^2 + 9x = 0\)
For this, do I just bring the 0 over and than add the 0 back in at the end as well, and than solve, or is it something different?

2. \(\displaystyle 3x^2 = 60\)
I'm really not sure where to go with this one, as it has no Bx and such...

3. \(\displaystyle (3x-4)^2 = 16\)

4. \(\displaystyle 4x^2 - 16 = 0\)

Also, for a couple problems I finished I'm just not sure if I got the correct answer. Would anyone be willing to just look them over for me and tell me whether I got them right or wrong?

A.
Problem: \(\displaystyle 4x^2 = 2x + 7\)

Answer: \(\displaystyle \frac {1 + \sqrt{29 }}{2}\) or \(\displaystyle \frac {1 - \sqrt{29 }}{2}\)

B.
Problem: \(\displaystyle x^2 -2x - 10 = 0\)

Answer: \(\displaystyle \frac {1 + \sqrt{11 }}{1}\) or \(\displaystyle \frac {1 - \sqrt{11 }}{1}\)

Any help or advice would be very much appreciated. Thank you!

 

[mmm4444bot]

1. \(\displaystyle x^2 + 9x = 0\)

do I just bring the 0 over Huh? Adding zero to both sides won't change anything.

Factor the lefthand side, and then use the Zero-Product Property

2. \(\displaystyle 3x^2 = 60\)

First, isolate x^2. Next, take the square root of both sides.

3. \(\displaystyle (3x-4)^2 = 16\)

Start by taking the square root of both sides.

4. \(\displaystyle 4x^2 - 16 = 0\)

The lefthand side is a Difference of Squares, so factor it.

Would anyone be willing to just look them over for me In a moment.

 

[mmm4444bot]

A. \(\displaystyle 4x^2 = 2x + 7\)

Answer: \(\displaystyle \frac {1 + \sqrt{29 }}{2}\) or \(\displaystyle \frac {1 - \sqrt{29 }}{2}\)

My numerators match yours, but I got a different denominator.

B. \(\displaystyle x^2 -2x - 10 = 0\)

Answer: \(\displaystyle \frac {1 + \sqrt{11 }}{1}\) or \(\displaystyle \frac {1 - \sqrt{11 }}{1}\)

These are both correct. However, they can be simplified.

If I were to ask you to write the number four on a piece of paper, would you write 4/1 ?