A2 U4.06

[cityracerstudio]

Alejandro.p.4
For the first problem I plugged solve (square root x•3)+(square root x+2)=3 and gives 1/6.
The second problem was solve 4(square root x•1)=x+3 which give x=1, 9.

[zxxBubBlZxxz]

David Ayala P.4
1. √(3-x)+√(x+2)=3. they intersect at 3 on the y-axis.
2. 4√x+1=x+3. The lines intersect at y=3 and y=15

[andrewramirez]

Andrew R p1.

1. √(3-x) + √(x+2) = 3 for this I got x= 2, 1
2. 4√x+1 = x+3 x= 6, -4sqr 2
Not to confident on these answers but while looking at the graph they intersected at one of these points.

[timmothyroberts]

Timmothy Roberts per 1

1.√3-x+√x+2=3
Answer x=-1 and x=2

2. 4√x+1=x+3
Answer x=5-4√2 and x=5+4√2

[Michael.Pineda]

Michael. P. 4

What I typed into Wolfram Alpha was "plot and solve (3-x)^1/2+(x+2)^1/2=3" and it worked out great. It gave me x=2, -1

My question was, is it best to divide 4 to both sides or to simply square the 4 along with √x when solving for the second problem?

[RyaAlexander]

Ryan D. 1
For the fourth practice problem is it neccesary to raise the equation to the second power twice, once in order to simplify and once after simplifying? 
When I used wolfram alpha for question 1, I said "graph, √3-x + √x+2=3" and x= approximately 2.3, for question 2, I said "plot, (4√x)+1=x+3".

[sergiojr]

Sergio Buenrostro Per. 1
The answer to number one is √(3-x)+√(x+2)=3 and the solution is x=-1 and x=2
Then, the answer to number two is 4√(x+1) = x+3 and the solution is x=5-4√2 and x=5+4√2
However, my question is that do you always have to multiply 1/2 of something to isolate the number outside of the square root?

[michaelromero]

Michael Romero P2
I plug  square root of (3-x) + square root of (x+2)= 3 into Wolfram Alpha
I am also confused on how you would plug in the second answer into Wolfram Alpha

[hectorcabrales]

Hector.C.Period 4
When it came to the second problem, I didn't know how to set up the equation after the regular steps.

Wolfram alpha I plotted but I'm not sure if it was right. Might of been extraneous because it gave me a negative solution

[Ace-qlyc]

Angel Carrillo Per4
1. My question is how do we write new equation, 6:00 of the video.
2.. √(3-x) + √(x+2) = 3
3.4√x+1 = x+3

[bv]

Brandon.BV.Period 2

Question: On the first long p problem with two radical expression equations, how did you exactly multiply the 1/2 to the 4 to equal 2 ?

Wolf: (3-x) + (x+2) = 3

[raullopez]

Raul Lopez
Period 2

My question is for the first one, do you we have to do additive inverse all the time?

(3-x)+(x+2)=3,

[EpicNightmarz]

Keith Sweet Per. 1
1. "Plot (square root symbol) 3 - x + (square root symbol) 2 + x the solution is x= -1
2. "Graph 4 (square root symbol) x+1"

I'm still confused about the 1/2 part

[robertomiranda]

Roberto M. Period 4
I'm not sure if I plugged in on the website correctly
1. √(3-x)+√(x+2)=3. On the graph 3 on the y-axis.
2. 4√x+1=x+3. On the graph y=3 and y=15

[chxngeless]

Chabuckets per. 2
What I typed into wolfram alpha for #1 was
- (3-x^1/2) (x+2^1/2) = 3 solve ..

My question would be why when we first squared the two square root equations we had to go through so much and even couldn't square the other side but the second time even though it was the exact same thing we were able to foil and square the other side also?

[briandelgadillo]

Brian D Period 1
What happens to the 2 in front of the √ x-1. For the first problem.

[brianflores]

Brian. F Per.4
I plugged in "graph (square root of 3-x)+(square root of x+2)=3" and I got 2, and -1.
How do you tell wolfram alpha to square root something because I have to type in "square root" in order for it to do it?

[albertoromero]

Alberto R. Per.1
For the first one how did you get 4 multiplied by 1/2?
1. For 3x is (-1, 2)
2. 4√x+1 = x+3 is x=6-4√2; x=6+4√2

[Richhhhhhard]

Richard M. Period 2 
1. √(3-x) + √(x+2) = 3 is how you plug it into Wolfram Alpha 
2. 4√x+1 = x+3 is how you plug it in and the answer is x= 6 +or - 4√2

[m]

Mathew C. Per. 2
I'm still confused with the first problem with the x/2 squared.