Question

Solve √3x+10 =x for x.

Answer 1

OK. You need to get all the Xes together in solving for x. If a square shows up, which may happen since there is a root, that can result in more than one answer.

Answer 2

What did you try to start and how far did you get?

Answer 3

I couldn't try because I didn't know how to start

Answer 4

\[x=10/(1-\sqrt{3})\]

Answer 5

AH. OK! Now, is that square root over the whole thing? All I see is it next ot the 3. So does it get the 10 also?

Answer 6

Yes, except for the =x part

Answer 7

OK. Then the fist step is to square both sides of the equation. This will let you get rid of the root.

Answer 8

that is what i think @e.mccormick it is next to the 3 except if not then the answer is different

Answer 9

3x+10=x^2

Answer 10

i took the x values to one side

Answer 11

@ballerina789 but if it is \[\sqrt{3x}\] then you have to square both sides

Answer 12

if it is \[\sqrt{3}x\] then all you have to do is take all the x components to one side

Answer 13

the √ goes over the whole equation. So when I square it it would equal: 3x+10=x^2

Answer 14

Yes.

Answer 15

Do you know where to go with that?

Answer 16

move everything to one side?

Answer 17

Yep. And a positive \(x^2\) is nice. So I would move the 3x+10 side.

Answer 18

x^2-3x+10=0

Answer 19

Now, can you factor that?

Answer 20

x^2-3x-x+10=0
?

Answer 21

Well, what you are looking for is something like:\[(x\pm\,?\,)(x\pm\,?\,)\]

Answer 22

(x-5)(x+2)

Answer 23

x=5
x=-2

Answer 24

correct?

Answer 25

Well, there is one other thing to do. You are almost there. Those arew correct, but you need to find x and verify it.

Answer 26

ok?

Answer 27

You see, because we started with \(\sqrt{3x+10}\) anything that makes the part under the root negative is NOT an answer.

Answer 28

So make sure that nothing you did would result in that being negative.

Answer 29

oh so the answer would only be 5

Answer 30

On a graph, only 5 works. In this case, -2 would mathematically work. But it does not work on a graph. Let me show you why:
https://www.desmos.com/calculator/a67kx3emql

Answer 31

ahhhh, ok. So the answer to the problem would be x=5

Answer 32

Yah. Now, the odd thing is caused by squaring it. Doing a square can add answers that do not work. When we squared we got: \(x^2-3x-10\). Let me see if I can add that to the graph.

Answer 33

Connection issues.... was going to show you one last thing. It will take a min. But I want you to get the point.

Answer 34

I added the \(x^2-3x-10\) to the graph.
https://www.desmos.com/calculator/pvjui3pw5b

Now, on the square, in orange, you are looking at the x intercepts. It shows two. However, the graph of the original equations is looking at the intersection.

The true test to see if the answers are right or wrong is to actually test them and do the math.

\(\sqrt{3(-2)+10}=-2\Rightarrow \sqrt{-6 +10}=-2 \Rightarrow 2=-2 \) which as you correctly saw, was false. In fact, anything under a root = to a negative is false, until you get to use complex numbers. So as long as the teacher has not said specifically, "Complex solutions," we assume real solutions and eliminate negatives like this.
\(\sqrt{3(5)+10}=5\Rightarrow \sqrt{15 +10}=5 \Rightarrow 5=5 \) Now that tests as true. So that is a real solution and useable.

The moral of the story is simple: Test answers for roots!

Answer 35

https://www.desmos.com/calculator/pvjui3pw5b

Now, on the square, in orange, you are looking at the x intercepts. It shows two. However, the graph of the original equations is looking at the intersection.

The true test to see if the answers are right or wrong is to actually test them and do the math.

\(\sqrt{3(-2)+10}=-2\Rightarrow \sqrt{-6 +10}=-2 \Rightarrow 2=-2 \) which as you correctly saw, was false.