Hi, I have a math project and I can not figure out these 5 problems, can you please help?
1. Solve x^2 + 2x + 9 = 0
a. x = -2 +/- 2i sqrt 2
b. x = -1 +/- 4i sqrt 2
c. x = -1 +/- 2i sqrt 2
d. x = -2 +/- 4i sqrt 2
2. Solve x^2 – 7x = –13
a. x = 7 +/- i sqrt 6 all over 2
b. x = -7 +/- i sqrt 6 all over 2
c. x = 7 +/- i sqrt 3 all over 2
d. x = -7 +/- i sqrt 3 all over 2
3.Solve –3x^2 – 4x – 4 = 0
a. x = 2 +/- 4i sqrt 2 all over 3
b. x = -2 +/- 4i sqrt 2 all over 3
c. x = 2 +/- 2i sqrt 2 all over 3
d. x = -2 +/- 2i sqrt 2 all over 3
4. Solve 5x2 = –30x – 65
a. x = -6 +/- 2i
b. x = -6 +/- 4i
c. x = -3 +/- 4i
d. x = -3 +/- 2i
5. Solve –3x2 + 30x – 90 = 0
a. x = 10 +/- 2i sqrt 5
b. x = 5 +/- 2i sqrt 5
c. x = 5 +/- i sqrt 5
d. x = 10 +/- i sqrt 5
Thank you, all help is helpful. It would be better to show me how you got the answers, if you give me the answers I will work through the problem to see how you got it if you do not show me how you did it.
Thank you very much!
Update:Thanks Don, the problem I have is reducing the square root like the sqrt of 48 is 2 sqrt of 18. I know that is not right, I just made it up but you get the point.
Answers & Comments
Verified answer
Cody,
I would just use the Quadratic Formula on all of these:
x = [-b ±√(b² - 4ac)) / 2a]
Plug in the values and find the correct solution. Easy!
In fact I'll do the first one for you.
1. Solve x² + 2x + 9 = 0
a=1, b=2, c=9
Pluggin in we have
x = [-2 ±√(2)²-4(1)(9) / 2(1)]
x = -1±2i√(2) -------------> Which is answer C.
You can do the rest!
:)
EDIT: Yes, I understand. Here's how I teach dealing with square roots. Use a factor tree to get all of the factors out and then pair them up.
For example, the one you gave: √(48)
So let's get all of the factors of 48 out:
48 ÷ 2 = 24
24 ÷ 2 = 12
12 ÷ 2 = 6
6 ÷ 2 = 3
So we have 2X2X2X2X3
So take out all the pairs: 2X2 and 2X2 and leave 3 (since it is alone)
Now you know that with a square root like √2X2 = √4 = 2
So for every two 2's, you take out one 2. OR,
(2)(2)√(3) = 4√(3) ----------> and that's the answer!
So the √(48) = 4√(3)
Does that help?
the gap between 2 factors relies on the pythagorean theorem. You calculate the finished upward push (difference between the y coordinates) and the run (difference between the x coordinates), and then build a suitable triangle such that the hypotenuse is the line between the two factors. -14 - (-2) = -12 (your "upward push") 12 - 9 = 3 (your "run") by way of pyth thm, (-12)^2 + 3^2 = d^2, the place d is the gap between the two factors. (-12)^2 + 3^2 = d^2 a hundred and forty four + 9 = d^2 153 = d^2 d = sqrt(153), or the sq. root of 153 = approximately 12.369. desire it helps.