Question

I need help arranging steps to an equation: Vx+3 - V2x-1=2
The Vs represent square root symbols.
These are my options.
1.)Simplify to obtain the final radical term
on one side of the equation.
2.)Raise both sides of the equation to
the power of 2.
3.)Apply the Zero Product Rule.
4.)Use the quadratic formula to find the
values of x.
5.)Simplify to get a quadratic equation.
6.)Raise both sides of the equation to
the power of 2 again.
I'd be grateful if someone could give me a short tutoring on how this exactly works.

Answer 1

I think that we can rewrite your equation as follows:
\[\large \sqrt {x + 3} = 2 + \sqrt {2x - 1} \]

Answer 2

Alright, I'm writing this down.

Answer 3

Raising the sides by 2 right? That's the first step?

Answer 4

yes!

Answer 5

step#2
we have to square both sides, so we get:

\[\large x + 3 = 4 + 2x - 1 + 4\sqrt {2x - 1} \]

Answer 6

Good, and then from the looks of it we have to simplify.

Answer 7

Oh, you already put it down.

Answer 8

step#3
we have to simplify that equation, as follows:

Answer 9

Into a quadratic equation yes? It's starting to come back to me.

Answer 10

\[\large \begin{gathered}
x + 3 - 4 - 2x + 1 = 4\sqrt {2x - 1} \hfill \\
- x = 4\sqrt {2x - 1} \hfill \\
\end{gathered} \]

Answer 11

step#4
we have to square both sides again:

Answer 12

\[\large {x^2} = 16\left( {2x - 1} \right)\]

Answer 13

After that there should be one final step.

Answer 14

step#5
I simplify that equation as follows:
\[\large {x^2} - 32x + 16 = 0\]

Answer 15

And that's to simplify.

Answer 16

yes! We can write:

Answer 17

step#6
\[\large \begin{gathered}
{x_1} = 16 + 4\sqrt {15} \hfill \\
{x_2} = 16 - 4\sqrt {15} \hfill \\
\end{gathered} \]
those are the solutions of the last equation

Answer 18

So the Zero Product rule is no concern here?

Answer 19

Or does that go with step 4?

Answer 20

namely I have used the quadratic formula

Answer 21

Ah, now I understand it.

Answer 22

I think I'm capable of starting my Algebra Test. Wish me luck.

Answer 23

we have to verify if those roots, are really the solutions of your irrational equation

Answer 24

those numbers are solutions to your irrational equation, if they satisfy the subsequent inequalities:
\[\large \begin{gathered}
x + 3 \geqslant 0 \hfill \\
2x - 1 \geqslant 0 \hfill \\
\end{gathered} \]

Answer 25

I passed!

Answer 26

I can assure that both those numbers verify those inequalities, so the solutions of your irrational equation, are:
\[\large \begin{gathered}
{x_1} = 16 + 4\sqrt {15} \hfill \\
{x_2} = 16 - 4\sqrt {15} \hfill \\
\end{gathered} \]

Answer 27

ok!

Answer 28

do you need more help?

Answer 29

Possibly in the future, as far as I'm concerned. For now I'm off to study on Polynomial Functions.

Answer 30

ok!

Answer 31

Thank you so much! I hope you have a great day!

Answer 32

Thank you! :)