Question

Medal+Fan for explanation. Solve this equation.
1+sqrt3x+3=sqrt7x+2

Answer 1

\[1+\sqrt{3x+3}=\sqrt{7x+2}\]

Answer 2

@ganeshie8

Answer 3

\[1+\sqrt{3x+3}=\sqrt{7x+2}\] squaring both sides,\[1+2\sqrt{3x+3}+(3x+3)=7x+2\]isloating the remaining root.\[2\sqrt{3x+3}=7x+2-(3x+3)-1\]

Answer 4

can you continue from here?
start from adding like terms on both sides.

Answer 5

Okay, let me see, one second. Could you check what I do?

Answer 6

sure sure...

Answer 7

can you please post any work you have here. if you have any questions, then please ask.

Answer 8

\[2\sqrt{3x+3}=7x+2-(3x+3)-1\]\[2\sqrt{3x+3}=7x+2-3x-3-1\]

Answer 9

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

Answer 10

Okay so then 2sqrt3x+3=4x+4?

Answer 11

why +4?

Answer 12

2+3-1?

Answer 13

it is negative 3.

Answer 14

I didn't distribute, my bad.

Answer 15

Som after that we divide both sides by 2.
\[\sqrt{3x+3}=2x-1\]

Answer 16

\[3x+3=4x^2-4x+1\]

Answer 17

\[0=4x^2-7x-2\]

Answer 18

Subtract out the left side, I'm following you.

Answer 19

lets test the discriminant. can you tell me what the dsicrimant is going to be?

Answer 20

You mean the factor correct?

Answer 21

no, I mean that you should (please)m find the discriminant for this.
if the discriminant is a perfect square, then the polynomial is factorable.

Answer 22

Well, it's not a perfect square.

Answer 23

\[(-7)^2-4(1)(-2)~~\Rightarrow~~49+8~~\Rightarrow~~57\]

Answer 24

you will have to use a quadratic formula thne.

Answer 25

can you fidn what x is?

Answer 26

by using the quadratic formula on 4x^2-7x-2

Answer 27

ohh, I did it incorrectly.

Answer 28

I mean my discriminant is,
\[(-7)^2-4(-4)(2)~~\Rightarrow~~49+32~~\Rightarrow~~81\]

Answer 29

it is a perfect square.

Answer 30

THANK GOODNESS! That's what I got after going back over it!

Answer 31

Okay, reminding the equation we have is,\[4x^2-7x+2=0\]

Answer 32

can you factor it?

Answer 33

One sec.

Answer 34

sure:)

Answer 35

Okay, this is quite confusing me. Can you assist?

Answer 36

fill in blanks.\[(4x~~~~~~~~~~~~)(x~~~~~~~~~~~~)\]

Answer 37

-7 +2? Doesn't it stay the same?

Answer 38

whay -7 and +2?

Answer 39

you need it to be,
\[4x^2-7x+2=0\]

Answer 40

I'm getting frustrated, factoring is the thing I'm quite worst at.

Answer 41

\[4x^2-7x+2=0\]\[(4x+1)(x-2)=0\]

Answer 42

can you tell me what x equals?

Answer 43

Yeah, I can do that.

Answer 44

4x+1=0 which means 4x=-1 OR x=-1/4

AND x-2=0 which means x=2

Answer 45

yes, x=-1/4 is one solution, and x=2.

Answer 46

Gr. Okay.

Answer 47

if you plug in -1/4 into square roots, do you get any negative numbers in the roots, or not?

Answer 48

No. Because a negative times a negative is never a negative right?

Answer 49

you need to make sure that when x=-/14 the square roots of your original equation are not negative

Answer 50

Oh, I see where you are. I'm sorry I'm so slow today. MAth is killing me. One moment.

Answer 51

take your time.

Answer 52

Uh, yeah they're negative. Duh.

Answer 53

The first isn't.

Answer 54

Oh wait. Neither is the second false alarm.

Answer 55

yes, none of the roots are negative when x=-1/4.

Answer 56

But it was important to check.... b/c we might have gootn some imaginary numbers.

Answer 57

So what does that mean?

Answer 58

Ohh, I see I see.

Answer 59

that x=-1/4 and x=2 are both fine solutions.

Answer 60

Thank you.. Do you mind one more?

Answer 61

okay, but I might have to leave at any second.

Answer 62

Okay.

Answer 63

Divide and Simplify if possible sqrt of 250x^16/sqrt2x

Answer 64

\[\Large \frac{\sqrt{250x^{16}}}{\sqrt{2x}}\]yes?

Answer 65

Exactly.

Answer 66

first, multiply times sqrtP2x{ on top and vbottom, to rationalize it.

Answer 67

sqrt{2x}

Answer 68

can you do this?

Answer 69

Nope. Listen if you're in a rush I can get someone else in /.\ don't wanna be a bother

Answer 70

I got to go, I am in arush... and that is nice of you:)
bye, and hope to see you some time later.