Question

Honors Algebra 2

Answer 1

\[1+ \sqrt {3x+3} = \sqrt {7x+2}\] I just need someone to walk me through the steps....

Answer 2

Okay, can you first subtract \(\large\color{black}{ \sqrt{3x+3} }\) from both sides?

Answer 3

okay....I can try.......

Answer 4

sure, go ahead;)

Answer 5

Or if the square roots scare you, you can square both sides

Answer 6

first we are subtracting as I said.

Answer 7

make the numbers more simple to work with, then go ahead and try to get x by itself.

Answer 8

Okay, haley, just subtract \(\large\color{black}{ \sqrt{3x+3} }\) from both sides, it shouldn't be a pain....

Answer 9

you are making it WAY more difficult than need be..

Answer 10

I'm really confused.

Answer 11

Idk which way it is easier, maybe you can raise both sides to the second power right away. it doesn't matter so much.

Answer 12

@SolomonZelman remember you can square both sides of an equation so you may want to keep the square roots on both sides of the equation.

Answer 13

not necessarily, but you can.

Answer 14

It just makes things easier sometimes! :)

Answer 15

it works either way.

Answer 16

okay, you guide the user then. we are all taking into 1 microphone.

Answer 17

I think I will try squaring because I can't remember very well how to subtract square roots....they confuse me

Answer 18

when you square both sides the the equation, the square roots cancel and you are left with
1+3x+3=7x+2

Answer 19

from there you want to get the x's to one side and the constants to the other

Answer 20

okay I was just confused on how to get to that, but that's easy now :)

Answer 21

yay!! glad we could help!

Answer 22

:) thanks!

Answer 23

\[1+ \sqrt {3x+3} = \sqrt {7x+2}\]subtract -1 from both sides.\[\sqrt{3x+3} =\sqrt{7x+2} -1\]now square both sides. Let \(a = \sqrt{7x+2}\) and \(b = -1\), then square this like you would any other term \(a^2 -2ab+b^2\)

Answer 24

what would be the point of moving the 1?

Answer 25

@Bryannaj12 Squaring both sides of the equation does not work that way.

Answer 26

Now I am just extremely confused...

Answer 27

uh yeah it does, how doesn't it?

Answer 28

\[(1+\sqrt{3x+3})^2 \ne 1+3x+3\]

Answer 29

yes it does, when you square a square root the square root cancels

Answer 30

\[(1+\sqrt{3x+3})^2 = 1 +2\sqrt{3x+3} +3x+3\]

Answer 31

oh goodness....I'm completely lost now

Answer 32

http://www.wolframalpha.com/input/?i=%281%2Bsqrt%283x%2B3%29%29%5E2

Answer 33

\[(1 +\sqrt{3x+3})^2 \implies (a^2 +b^2) = a^2 +2ab +b^2\]

Answer 34

Always remember that. You can not easily get rid of a square root in that method.

Answer 35

so it would be \(1+ \sqrt {3x+3}+3x+3\)

Answer 36

whoopsies! forgot the 2..

Answer 37

Yes, which is reduced to \(3x +2\sqrt{3x+x} +4\)

Answer 38

uhmmm how did you get the x inside the sqrt? I thought it was +3?

Answer 39

Typo!

Answer 40

Okay.

Answer 41

ahh, i forgot to foil.

Answer 42

\[3x +2\sqrt{3x+3} +4\]

Answer 43

so then it would equal
\(3x +2\sqrt{3x+3} +4=7x+2\)

Answer 44

oh! give me a second to figure the rest out and could you just check my answer?

Answer 45

So now you have: \(3x+2\sqrt{3x+3}+4 = 7x+2\) Correct. \(\checkmark\)

Answer 46

Okay I'm stuck....I got as far as \(\sqrt {3x+3}=2x-1\) correct?

Answer 47

then I got \(3x+3=4x^2-4x+1\)

Answer 48

That's correct.

Answer 49

then \(4x^2-7x-2=0\)

Answer 50

And do you know how to solve these types of quadratics?

Answer 51

That can't really factor, can it? So I have to use the quadratic formula...

Answer 52

There are a couple different methods for solving, completing the square, or factoring.

Answer 53

I personally avoid using the quadratic formula.

Answer 54

okay I know how to complete the square.... :) one sec

Answer 55

Would you like me to show you both methods?

Answer 56

Oh alright.

Answer 57

completing the square is my favorite method, although a lot of people avoid uing it due to its complexities...

Answer 58

I think I prefer (in this case because I get 4x^2-7x+12.25) because I don't like to factor decimals.

Answer 59

\[x=\frac{ 7 \pm \sqrt {49-4(4)(-2)} }{ 8 }\]

Answer 60

Oh you wont get decimals by completing the square Let me show you.

Answer 61

For this one I think quadratic formula will be easier for me....

\(x=\frac{ 7 \pm \sqrt {49+32} }{ 8 }\)

\(x=\frac{ 7 \pm \sqrt {81} }{ 8 }\)
\(x=\frac{ 7 \pm 9 }{ 8 }\)

Answer 62

I got x=-1/4 and x=2

Answer 63

@Jhannybean Is doing it all correct. :)

Answer 64

You are in good hands @haleyelizabeth2017

Answer 65

\[4x^2-7x-2=0\]First group your x terms. \[(4x^2 -7x) -2=0\]Now factor out a 4.\[4\left(x^2-\frac{7}{4}x\right) -2=0\]Find yur new \(c\) value \(c=\left(\frac{b}{2}\right)^2\)And remember, because we factored out a 4, our new c value has to be multiplied to it.\[4\left(x^2 -\frac{7}{4}x +\frac{49}{16}\right)-2-\frac{49}{16}=0\]\[4\left(x-\frac{7}{8}\right)^2 -\frac{81}{16}=0\]\[4\left(x-\frac{7}{8}\right)^2 = \frac{81}{16}\]\[\left(x-\frac{7}{8}\right)^2 = \frac{81}{64}\]

Answer 66

yep, I got the same answer as well.

Answer 67

\[x= \pm \frac{9}{8} +\frac{7}{8} \implies 2 ~,~ -\frac{1}{4}\] :)

Answer 68

okay great! :D

Answer 69

Thank you!

Answer 70

no problem :)

Answer 71

I goofed on one step. \[4\left(x^2 -\frac{7}{4}x +\frac{49}{64}\right)-2-\frac{49}{16}=0\]

Answer 72

Okay... :)