I have the answers to these problems, i just need to know HOW to get them?

Question

By using the quadratic formula, solve:
1. x^2-2x-7=0
2. 2/3x^2+5/6x+1/2=0


Please help me out.. I have my final tomorrow and I am struggling with these problems. I have the answers, also, I just didn't know how to type them out because they have the square root sign as well and the "plus or minus" sign... Thank you so much for your help in advance.

Answer 1

First... the symbols
√∛∜≤≥±
are all in MS Word. Click "Insert", Click "Symbol", Select "(normal text)" Format. Select "Mathematical Symbols" Subject. They're all there.. including exponents and subscripts. Some set-theoretical notation doesn't cut and paste here. And they cut and paste here.

1. x^2-2x-7=0
The Quadratic Formula
x = {-(b) ±√[(b)² - 4(a)(c)]}/[2(a)]
Substituting
x = {-(-2) ±√[(-2)² - 4(1)(-7)]}/[2(1)]
x = {2 ±√[4 - -28]}/2
x = {2 ±√[32]}/2
x = {2 ±4√[2]}/2
x = 1+ 2√2
OR x =1- 2√2


2. 2/3x^2+5/6x+1/2=0
I hope this is (2/3)x² + (5/6)x + 1/2=0
If so, then mujltiply both sides by 6
4x² + 5x + 3 = 0
The Quadratic Formula
x = {-(b) ±√[(b)² - 4(a)(c)]}/[2(a)]
Substituting
x = {-(5) ±√[(5)² - 4(4)(3)]}/[2(4)]
x = {-5 ±√[25 - 48]}/8
x = {-5 ±√[-23]}/8
x = {-5 ± i√[23]}/8
I don't know if -5/8 ± i√[23]/8 is any better..

Note, in MS Word you can italicize the i.

If, however, if 2 is 2/(3x^2)+5/(6x)+1/2=0
Then multiply both sides by 6x² and you get
4 + 5x + 3x² = 0
Substituting
x = {-(5) ±√[(5)² - 4(3)(4)]}/[2(3)]
x = {-5 ±√[25 - 48]}/6
x = {-5 ±√[-23]}/6
x = {-5 ± i√[23]}/6

Answer 2

think of your equations in this form:

ax^2 + bx + c

so...

x^2 - 2x - 7
For number 1. A = 1 (there is only 1 x^2)
B = -2 (minus 2x)
C = -7 (minus 7)

then you plug a, b, and c into the quadratic formula

which is:

-b plus or minus the square root (b^2 - 4(a)(c))/2a

for number 2
A = 2/3
B = 5/6
C = 1/2

Answer 3

Question 1
x = [ 2 ± √(4 + 28) ] / 2
x = [ 2 ± √32 ] / 2
x = [ 2 ± 4√2 ] / 2
x = 1 ± 2√2

Question 2
4x² + 5x + 3 = 0
x = [ - 5 ± √(25 - 48) ] / 8
x = [ - 5 ± √(- 23) ] / 8
x = [ - 5 ± i √23 ] / 8

Answer 4

they are both in the form

ax^2 + bx +c = 0

using the quadratic formula

x = (-b +- sqrt(b^2 - 4ac))/2a

just put in the values

1. x= (2 +- sqrt(4-4*1*-7))/2

x= (2 +- sqrt (32)) /2

hope thats the answer you have

just follow the same steps for the 2nd question

Answer 5

1.
x^2-2x-7=0
compare with ax^2+bx+c=0
a=1, b=-2, c=-7
x={-b+sqrt(b^2-4ac)}/2a and {-b-sqrt(b^2-4ac)}/2a
x={2+sqrt(4+28)}/2 and {2-sqrt(4+28)}/2
x={2+4sqrt(2)}/2 and {2-4sqrt(2)}/2
x=1+2sqrt(2) and 1-2sqrt(2)

2. solve as above