Solving radical equations | Exponent expressions and equations | Algebra I | Khan Academy

Solving Radical Equations

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We're asked to
solve the equation,

3 plus the principal square root
of 5x plus 6 is equal to 12.

And so the general strategy
to solve this type of equation

is to isolate the radical sign
on one side of the equation

and then you can square
it to essentially get

the radical sign to go away.

But you have to be
very careful there

because when you
square radical signs

you actually lose the
information that you were

taking the principal
square root.

Not the negative square root
or not the plus or minus square

root.

You are only taking the
positive square root.

And so when we get
our final answer,

we do have to
check and make sure

that it gels with taking
the principal square root.

So let's try.

Let's see what
I'm talking about.

So the first thing
I want to do is

I want to isolate this on
one side of the equation.

And the best way to isolate
that is to get rid of this 3.

And the best way
to get rid of the 3

is to subtract 3 from
the left-hand side.

And of course, if I do
it on the left-hand side

I also have to do it
on the right-hand side.

Otherwise, I would
lose the ability

to say that they're equal.

And so the left-hand
side right over here

simplifies to the principal
square root of 5x plus 6.

And this is equal to 12 minus 3.

This is equal to 9.

And now, we can square both
sides of this equation.

So we could square the principal
square root of 5x plus 6

and we can square 9.

When you do this-- when you
square this, you get 5x plus 6.

If you square the square
root of 5x plus 6,

you're going to get 5x plus 6.

And this is where we actually
lost some information

because we would
have also gotten this

if we squared the negative
square root of 5x plus 6.

And so that's why we have to be
careful with the answers we get

and actually make sure it works
when the original equation

was the principal square root.

So we get 5x plus 6
on the left-hand side.

And on the right-hand
side we get 81.

And now, this is just a
straight up linear equation.

We want to isolate the x terms.

Let's subtract 6
from both sides.

On the left-hand side, we have
5x and on the right-hand side,

we have 75.

And then we can divide
both sides by 5.

We get x is equal to--
let's see, it's 15, right?

5 times 10 is 50.

5 times 5 is 25 gives you 75.

So we get x is
equal to 15, but we

need to make sure
that this actually

works for our original equation.

Maybe this would
have worked if this

was the negative square root.

So we need to make
sure it actually

works for the
positive square root,

for the principal square root.

So let's apply it to
our original equation.

So we get 3 plus the principal
square root of 5 times 15.

So 75 plus 6.

So I just took 5
times 15 over here.

I put our solution in.

It should be equal to 12.

Or we get 3 plus square
root of 75 plus 6

is 81 needs to be equal to 12.

And this is the principal
root of 81 so it's positive 9.

So it's 3 plus 9 needs
to be equal to 12,

which is absolutely true.

So we can feel pretty
good about this answer.