Solving Radical Equations - I

demha · 2015-03-31 06:02:30

Hey guys I'm back! I need some help understanding how to solve some equations and checking the ones I solved already to see if they are right. For the ones I need help solving, I would prefer to be helped understand how to get the answer rather than just being given the answer. As usual I only have about 20 math problems so I'l be posting 5 of them at a time. Here is the first part:

1. [SQRT(x - 3)] - 7=0

A -46
B 10
C 52
D 4
E {4,10}
F none of the above

My Answer: C

2. 2[SQRT(w + 4)]=5

A 1
B 9/4
C -1/2
D 17/2
E -9/4
F none of the above

not sure how to solve

3. [SQRT(x)]=-2

A 0
B 4
C-4
D-2
E no solution
F all real numbers

My Answer: E

4. [SQRT(2x2 - 1)]=x

A[SQRT(1/3)]
B {-1,1}
C {-1/2,1}
D 1
E 0
F none of the above

not sure how to solve

5. [SQRT(2x^2 + 5x + 6)]=x

A no solution
B {-2,-3}
C {2,3}
D -2
E 3
F 0

not sure how to solve

Bob · 2015-03-31 06:15:23

hi demha

Q!. Correct.

Q2. Divide each side by 2, square and subtract 4.

Q3. If you square both sides you'll see this has a solution. (I'm assuming a square root could be + or - )

Q4. Simplify the bracket. Do you see that square root amongst the answers?

Q5. You could square both sides, re-arrange and re-factorise. But what I did was try the possible solutions to see if any fit. I got an answer straight away.

Bob

ROA · 2015-04-01 00:02:46

Hi Bob,

Q4 refers pls.

Do you see SQRT among the answers might be tempting.

Pls check what follows SQRT in the answer.

ROA

Bob · 2015-04-01 03:08:58

hi ROA,

Thanks. I have looked again and think I may have mis-interpreted the question.

4. [SQRT(2x2 - 1)]=x
A[SQRT(1/3)]
B {-1,1}
C {-1/2,1}
D 1
E 0
F none of the above

I was reading it as

in which case x = √3

That's what happens when posters don't use LaTex.

If the question is

then a good starting point is the square both sides:

Bob

demha · 2015-04-01 03:31:06

So sorry about the misunderstanding! Q4 is indeed 2x^2. I see the latex page and I'll be going with that from now on.

2.

A 1
B 9/4
C -1/2
D 17/2
E -9/4
F none of the above

3.

A 0
B 4
C-4
D-2
E no solution
F all real numbers

My Answer: E

4.

A
B {-1,1}
C {-1/2,1}
D 1
E 0
F none of the above

5.

A no solution
B {-2,-3}
C {2,3}
D -2
E 3
F 0

NOTE:
Currently editing this post to make sure that the latex comes out right.

Last edited by demha (2015-04-01 03:50:10)

demha · 2015-04-01 04:03:48

Q2.

Divide both sides by 2:

How should I carry on from here?

Q3.
Well, I read my lesson and it says "no positive square root can ever possibly equal a negative number." But if we could go on with it I would say the answer is 'B.'

Last edited by demha (2015-04-01 04:35:35)

Bob · 2015-04-01 05:30:37

hi

Q3. B is correct.

"no positive square root can ever possibly equal a negative number."

This is an odd statement. The notes might as well have said "No positive number can ever possibly equal a negative number." Squaring the left and right sides removes the negative.

Q2. Not quite.

This becomes

So those 2s on the left cancel and you can square both sides.

Thanks for using LaTex. That makes things a lot clearer.

My hints for Q4 and Q5 are unchanged.

Bob

phrontister · 2015-04-01 10:33:30

demha wrote:

Q4 is indeed 2x^2.

An alternative is 2x² that uses the superscript '²' from the group of 'Useful symbols' given at the top of the page...but Latex is clearest.

Math Is Fun Forum

#1 2015-03-31 06:02:30

Solving Radical Equations - I

#2 2015-03-31 06:15:23

Re: Solving Radical Equations - I

#3 2015-04-01 00:02:46

Re: Solving Radical Equations - I

#4 2015-04-01 03:08:58

Re: Solving Radical Equations - I

#5 2015-04-01 03:31:06

Re: Solving Radical Equations - I

#6 2015-04-01 04:03:48

Re: Solving Radical Equations - I

#7 2015-04-01 05:30:37

Re: Solving Radical Equations - I

#8 2015-04-01 10:33:30

Re: Solving Radical Equations - I

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