Question

\sqrt{x^{2}+3}+\sqrt{x+3}=5

Answer 1

\(\sqrt{x^{2}+3}+\sqrt{x+3}=5\)

Answer 2

yes

Answer 3

but I need doing post.Can you help me

Answer 4

Square both sides as a first step.

Answer 5

yes I square

Answer 6

What do you get?

Answer 7

You can not solve for me?

Answer 8

So I tried every possible way

Answer 9

Can you square the left side?

Answer 10

YES

Answer 11

but it's 25

Answer 12

I use application
\[2\sqrt{AB}=C-A-B\]

Answer 13

It might be easier if we take one radical to the other side:
\[
\sqrt{x^{2}+3}+\sqrt{x+3}=5 \\
\sqrt{x^{2}+3} = 5- \sqrt{x+3} \\
\text{Square both sides:} \\
x^2 + 3 = 25 + x + 3 - 10\sqrt{x+3} \\
x^2 + 3 - 25 - x - 3 = - 10\sqrt{x+3} \\
x^2 - x - 25 = - 10\sqrt{x+3} \\
\text{Square both sides again:}
\]Can you do that part?

Answer 14

yes I worked to do so still can not calculate when I put t

Answer 15

Because if it's a 4 set t hat do I do for the next star?

Answer 16

?

Answer 17

\[
x^2 - x - 25 = - 10\sqrt{x+3} \\
\text{Square both sides again:} \\
x^4 + x^2 + 625 - 2x^3 + 50x - 50x^2 = 100x + 300 \\
x^4 - 2x^3 - 49x^2 - 50x +325 = 0
\]

Answer 18

Well I was out there, but when they do not get put t

Answer 19

Are you allowed to use graphing calculators? If allowed we may not even need these steps and just plot the original equation.

Answer 20

Yeah but my teacher began to explain and not allowed to use computers or rounding

Answer 21

Graphing calculators allowed or not allowed?

Answer 22

not allowed

Answer 23

What chapter is this problem from? Numerical Analysis? Give some background so I get an idea what is allowed here.

Answer 24

?????

Answer 25

Well this is all based solution to find x such that this equation based experience satisfying x = 5, is not allowed to rounding or computer use

Answer 26

Yes we have to solve for x. But we have a 4th degree equation. We can find solutions by trial and error. Is that allowed?
We can try x = 1, x = 2, x = 3, etc, until we find two values where the function changes sign. Then there must be a root between those two numbers.Then we refine it further and find a solution to the first decimal place , then second decimal place, etc. It will take a lot of time and a lot of trial and error.

Answer 27

Is it because my math teacher for only 2.11 Acceptance very odd but I began to find I do not know how to prove

Answer 28

Oh, wait a minute! They gave you x = 2.11 and asked you to prove that it is a solution for the given equation?

Answer 29

Are you asked to prove x = 2.11 is a solution?

Answer 30

No, I press the computer to try to find the test before, but no way to prove it with this experiment or solve all the equations

Answer 31

Shift +solve is Answer but I don't know solve problems

Answer 32

Solving this problem requires one of the following:
!. Graphing Calculator
2. Online graphing calculator
3. Trial and error method

If the first two are not allowed I can only think of the third choice to solve this problem.

Answer 33

http://www.wolframalpha.com/input/?i=sqrt%28x^2%2B3%29+%2B+sqrt%28x%2B3%29+%3D+5

Answer 34

Thank you

Answer 35

I will go to bed now good day for you.Thank you to help me

Answer 36

You are welcome.

Answer 37

Trial and error method:
\[
f(x) = \sqrt{x^{2}+3}+\sqrt{x+3} =5 \\
f(0) = \sqrt{0^{2}+3}+\sqrt{0+3} = 3.46 \\
f(1) = \sqrt{1^{2}+3}+\sqrt{1+3} = 4 \\
f(2) = \sqrt{2^{2}+3}+\sqrt{2+3} = 4.88 \\
\text{Answer getting closer to 5...}
\]

Answer 38

\[
f(x) = \sqrt{x^{2}+3}+\sqrt{x+3} =5 \\
f(2.1) = \sqrt{2.1^{2}+3}+\sqrt{2.1+3} = 4.98 \\
f(2.11) = \sqrt{2.11^{2}+3}+\sqrt{2.11+3} = 4.99 \\
f(2.12) = \sqrt{2.12^{2}+3}+\sqrt{2.12+3} = 5.0003 \\
x \approx 2.12
\]

Answer 39

thanks