1. ## algebra equation

question 1: ^ = power of
if 3x^2-2x+7=0 then (x-1/3)^2= ?
answer : -20/9 HOW DID THEY GET THAT? I DONT UNDERSTAND IT.

question 2:
the graph of which of the following equations is a straight line parallel to the graph of y = 2x
answer: 2x-y=4 HOW THEY GET THAT?

question 3:
An apartment building contains 12 units consisting of 1 and 2 bedroom apartments that rent for $360 and$450 per month, respectively. When all the units are rented, the total monthly rental is $4950. What is the number of 2 bedroom houses? answer: 7 I thought I was doing this problem right with multipling 450 with each choice and then subtracting that from 4950 then dividing that from 360 and thought the right answer would be the difference I needed to make 12 units but that didnt work so I dont know how they got that answer. question 4: Have one square with a 125 written inside it. another little square with a 5 written in it. If the 2 square regions in the figures below have the respective areas indicated in the square yards, how many yards of fencing are needed to enclose the two regions? answer:24(5 root sign over it) if that makes any sense to you. please help me. 2. For Qn 3 Solve the simulataneous equation listed below : Let the number of 1 bedrooms be x Let the number of 2 bedrooms be y x + y = 12 ---------- Eqn 1 360x + 450y = 4950 ------ Eqn 2 You will get x = 5 and y = 7 3. can u show me how you got x = 5 y = 7 because I keep doing the math and getting it wrong. I use that equation you gave me and try to solve for x and y but I just get stummed. Appricate it. 4. I can't honestly say I agree with the answer you have for Question 4. I'll show you how I solved it and see if that makes any sense to you. You have one square with an area of 125 square yards, which means that: $A_1 = s_1^2 = 125 \ yds^2$ $s = \sqrt{125} \ yds$ We also have another square with an area of 5 square yards, which means that: $A_2 = s_2^2 = 5 \ yds^2$ $s = \sqrt{5} \ yds$ We also know that the diagram looks a lot like: So, the perimeter of the 125 square yard square is: $P_1 = 4\sqrt{125}$ And the perimeter of the 5 square yard square is: $P_2 = 4\sqrt{5}$ We know that: $125 = 5(25)$ $\sqrt{125} = \sqrt{5(25)} = \sqrt{25}\sqrt{5} = 5\sqrt{5}$ Therefore: $P_1 = 4(5\sqrt{5}) = 20\sqrt{5}$ So, the total to enclose the squares is: $T = P_1 + P_2$ $= 20\sqrt{5} + 4\sqrt{5}$ $= 24\sqrt{5}$ There you go. I hope that's what you meant. 5. Originally Posted by girl in distress question 1: ^ = power of if 3x^2-2x+7=0 then (x-1/3)^2= ? answer : -20/9 HOW DID THEY GET THAT? I DONT UNDERSTAND IT. question 2: the graph of which of the following equations is a straight line parallel to the graph of y = 2x answer: 2x-y=4 HOW THEY GET THAT? question 3: An apartment building contains 12 units consisting of 1 and 2 bedroom apartments that rent for$360 and $450 per month, respectively. When all the units are rented, the total monthly rental is$4950. What is the number of 2 bedroom houses?

answer: 7 I thought I was doing this problem right with multipling 450
with each choice and then subtracting that from 4950 then
dividing that from 360 and thought the right answer would
be the difference I needed to make 12 units
but that didnt work so I dont know how they got that

question 4:
Have one square with a 125 written inside it. another little square with a 5 written in it. If the 2 square regions in the figures below have the respective areas indicated in the square yards, how many yards of fencing are needed to enclose the two regions?

For 1. Notice that this is a quadratic.

Using the discriminant

$\Delta = b^2 - 4ac$

$= (-2)^2 - 4(3)(7)$

$= 4 - 84$

$= -80 < 0$.

So the solutions are non-real (i.e. complex conjugates).

Now solve for x...

$x = \frac{-b \pm \sqrt{\Delta}}{2a}$

$= \frac{2 \pm \sqrt{-80}}{2(3)}$

$= \frac{2 \pm 4\sqrt{5}i}{6}$

$= \frac{1}{3} + \frac{2\sqrt{5}i}{3}$ or $= \frac{1}{3} - \frac{2\sqrt{5}i}{3}$.

Now substitute these values into $\left(x - \frac{1}{3}\right)^2$ and see what you get.

6. Sure. Let me continue from where i started.

For Qn 3 Solve the simulataneous equation listed below :

Let the number of 1 bedrooms be x
Let the number of 2 bedrooms be y

x + y = 12 ---------- Eqn 1
360x + 450y = 4950 ------ Eqn 2

from Eqn 1 make x the subject : x = 12 - y ------ Eqn 3
Substitute Eqn 3 into Eqn 2

360(12-y)+450y = 4950
4320-360y+450y=4950
90y=630
y=7

substitute y=7 into eqn 1
x+7=12
x=5

Hope it helps.

Originally Posted by girl in distress
can u show me how you got

x = 5 y = 7 because I keep doing the math and getting it wrong. I use that equation you gave me and try to solve for x and y but I just get stummed. Appricate it.

7. ## Completing the Square

Hello girl in distress

Originally Posted by girl in distress
question 1: ^ = power of
if 3x^2-2x+7=0 then (x-1/3)^2= ?
answer : -20/9 HOW DID THEY GET THAT? I DONT UNDERSTAND IT.

question 2:
the graph of which of the following equations is a straight line parallel to the graph of y = 2x
answer: 2x-y=4 HOW THEY GET THAT?
Question 1

The trick you need to understand here is called Completing the Square, and it goes like this.

First, a 'math-y' word: coefficient, which means the number that's in front of the
$x$ or $x^2$ (along with its minus sign if it has one). So in your expression, the coefficient of $x^2$ is $3$, and the coefficient of $x$ is $-2$.

So, this is how you complete the square (it's a bit complicated!):

Step 1: Take the coefficient of
$x^2$ outside a bracket, dividing all the other cofficients to make it correct. Like this:

$3x^2-2x+7=3(x^2-\frac{2}{3}x+\frac{7}{3})$

(Do you see how the
$-2$ and the $+7$ have been divided by $3$, making $-\frac{2}{3}$ and $\frac{7}{3}$?)

Step 2: Inside the brackets you just made write another pair of brackets that contain:
$x$ and half the coefficient of $x$ in the previous line. That's:

$3((x-\frac{1}{3})...)$

(Do you see what I've done? The coefficient of
$x$ in the previous line was $-\frac{2}{3}$, and I've halved it to get $-\frac{1}{3}$.)

Step 3: write a power
$^2$ after this new bracket. So that is simply:

$3((x-\frac{1}{3})^2...)$

Step 4: Subtract the square of the number you just wrote in the new bracket. That number was
$-\frac{1}{3}$, and the square of this number is $\frac{1}{9}$. So:

$3((x-\frac{1}{3})^2-\frac{1}{9}...)$

Step 5: Put back the final number that you had when you first created the original bracket. That was
$+\frac{7}{3}$. So:

$3((x-\frac{1}{3})^2-\frac{1}{9}+\frac{7}{3})$

Step 6: Simplify the fractions. That's
$-\frac{1}{9}+\frac{7}{3}$, which makes $\frac{20}{9}$. So we get:

$3((x-\frac{1}{3})^2+\frac{20}{9})$

Putting all this together into your equation, we get:

$3x^2-2x+7=0$

$\implies 3(x^2-\frac{2}{3}x+\frac{7}{3})=0$

(You can now divide both sides of the equation by 3, and leave out the outer brackets, if you like. If it's just an expression, not an equation, you can't do this; you have to leave the 3 there.)

$\implies x^2-\frac{2}{3}x+\frac{7}{3}=0$

$\implies (x-\frac{1}{3})^2+\frac{20}{9}=0$

Re-arrange the equation:

$\implies (x-\frac{1}{3})^2=-\frac{20}{9}$

Question 2

Is much easier. It's all to do with gradients. Three things you need to know:

1 How to re-arrange equations like
$ax+by+c=0$ into something like $y=...$

2 The gradient of the line whose equation is
$y=mx+c$ is $m$, the coefficient of $x$.

3 Parallel lines have equal gradients.

$y=2x$ is $2$.

Re-arrange
$2x-y=4$:

$y=2x-4$

The gradient of this line is also
$2$, so the lines are parallel.

OK?

8. girl in distress, thank you so much for posting this! you have no idea how happy I am that I found this. For anyone who doesn't know, these are questions from the collegeboard ACCUPLACER exam sample questions. I have been trying to find help on these forever.

thank you thank you thank you

and thank you to everyone who helped!

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# If the two square region in the figure below have the respective areas indicated in square yards

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