Who is really good in algebra 2?

Question

I have struggled with some concepts in algebra. The school that taught me algebra 1 did a poor job in teaching, and I literally struggled with it. I am currently having trouble with completing the square. Here are three examples that I have on my homework.
1. n^2 + 16n -7 = 0
2. 3x^2 -5x + 2 = 0
3. 2d^2 - 10d + 5 = 0

I just want to know how to do this. I get really upset when I don't understand this math.

Answer 1

I did have a tough time with this stuff for myself. I found a website (in the source area below) and it may help you understand the procedure a little better if you work with it.

Answer 2

Completing the square just means taking the first two terms and creating a perfect square out of it. Think about the pattern of the perfect square:

(x + a)^2 = x^2 + 2ax + a^2

(x - a)^2 = x^2 - 2ax + a^2

First off, you'll notice that the sign of the linear term tells you which one you're dealing with. If the middle term is negative, it's (x - a)^2. If positive, it's (x + a)^2. Also, note that in both cases, the coefficient of the linear term is twice the square root of the constant term.

So:

1) n^2 + 16n - 7 = 0

We're dealing with an (x + a)^2 situation here, because the 16n is positive. So, from the pattern,

n^2 + 2an + a^2 = 0
n^2 + 16n + a^2 = 0

Clearly, 2a is 16, so

a = 8

Now we need to massage the equation into the proper form. The perfect square would be n^2 + 16n + 64, but we have n^2 + 16n - 7. What do we need to add to the equation to turn -7 into +64? Clearly, it's 71. So:

n^2 + 16n - 7 = 0

n^2 + 16n - 7 + 71 = 0 + 71

n^2 + 16n + 64 = 71

(n + 8)^2 = 71

Now we can solve it:

n + 8 = +/- sqrt(71)

(sqrt() = square root)

n = -8 +/- sqrt(71)

2) For this one, we need to get tricky. Both of the patterns involve a unary quadratic term - i.e. the coefficient of the squared term must be 1. So, we start by doing that:

3x^2 - 5x + 2 = 0

x^2 - 5/3 x + 2/3 = 0

Now: take 5/3 and cut it in half to get 5/6. Square that and get 25/36. So the perfect square we want is

x^2 - 5/3 x + 25/36 = (x - 5/6)^2

We need to add something to turn 2/3 into 25/36. 2/3 = 24/36, so we need to add 1/36.

x^2 - 5/3x + 2/3 + 1/36 = 1/36

x^2 - 5/3 x + 25/36 = 1/36

(x - 5/6)^2 = 1/36

x - 5/6 = +/- 1/6

x = 5/6 +/- 1/6

So, x = 1 or x = 4/6 = 2/3

Work the third one the same way as #2.

Answer 3

Step 1: Divide thru to make the coeffiecient of the squared term be 1. So in problem #2 you would divide thru by 3, and problem #3 by 2 (e.g. #3 becomes d^2 - 5d + 5/2 = 0
Step 2: Take the constant term to the other side of the equal sign (i.e. d^2 - 5d = -5/2)
Step 3: take 1/2 the coefficient of the term without the squared in it, and add to both sides. In this case you take 1/2 of -5 (since -5d is the term without the squared). Thus you get d^2 -5d + 25/4 = 5/2 + 25/4
Step 4: The equation now becomes (d - 5/2)^2 = 35/4
Step 5: Solve for d by taking the square root of the right hand side, then carrying the constant term over form the left hand side (i.e. d = [sqrt (35/4) + 5/2])

Hope that helps!! It can be kinda tricky

Answer 4

For increasing cubic binomials the overall formula is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^a million + 3*a^a million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^a million + 3*x^a million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D

Answer 5

I am greek so for words i dont know i will use the greek ones. sorry

if you have a 2nd grade equation and it is like the following

ax^2 + bx + c=0 then first you have to find the "Diakrinousa"

D= b^2-4ac
if D>0 then there are 2 solutionsthen
x1=(-b+D^1/2)/2a
x2=(-b-D^1/2)/2a
if D=0 there is one (double) solution
then x1=x2=(-b)/2a
if D<0 there is no solution in real numbers.

then x1=(-b+D^1/2)/2a
x2=(-b-D^1/2)/2a

Answer 6

for numbr 2.. its 3x^2-3x-2x+2= 3x-2
=<3x-2>

for da oda two probs are u sure u copied da ques rite.. if u did then u gotta use that quadratic formula itz like x=b-sqrt4ab-c som shyt...

Answer 7

im pretty sure number one is 16n^2. im not sure my algabra level is up that far yet but im pretty good at it. and im sorry if it was the wrong answer. im not sure about the others though

Answer 8

YOU ARE GOING TO NEED THE QUADRATIC FORMULA TO SOLVE THESE PROBLEMS. THE ANSWERS ARE GOING TO HAVE SQUARE ROOTS AND FRACTIONS IN THEM. THE EQUATION IS VERY DIFFICULT TO TYPE, SO PLEASE LOOK IT UP IN YOUR MATH BOOK. THE ANSWER FOR #2 IS X=1 AND X= 2/3 IT'S EASY WHEN YOU USE THE FORMULA EQUATION. GOOD LUCK.

Answer 9

1. -23
2. -0.5
3. 5/6

Answer 10

1. -23
2. -0.5
3. 5/6