Can anyone help me to set this up?

solve: (√2m+15)=m

everything on the left side of the equation is under the sq. root.

thanks

2. ## Re: Radical Equation help

Take the square of both sides:
$\displaystyle \sqrt{2m+15}=m$
$\displaystyle \Leftrightarrow \left(\sqrt{2m+15}\right)^2=m^2$

(Don't forget to determine the 'existence conditions' because $\displaystyle 2m+15$ has to be $\displaystyle >0$ or $\displaystyle =$ 0)

3. ## Re: Radical Equation help

ok so would the next step be -m^2+2m+15 >=0 ?

Thanks

4. ## Re: Radical Equation help

Yes, that's correct, but you have understand me wrong, I didn't say that:
$\displaystyle -m^2+2m+15>=0$
You just can wright:
$\displaystyle -m^2+2m+15=0$

I mean, that after you've calculated the solutions, you have to assure that $\displaystyle m>=\frac{-15}{2}$(conditon 1) and $\displaystyle m>0$ (condition 2) summarized $\displaystyle m>0$, because you've to assure you don't get a negative number under the square root.

5. ## Re: Radical Equation help

I find it easier to take $\displaystyle a$ as being positive since IMO it makes factorising it easier. To do this you can move everything over to the other side (and not forgetting to change the sign) :
$\displaystyle m^2-2m-15=0$. This quadratic does factor but there is nothing wrong with the formula if you prefer it.

Because you squared both sides of the equation you may have introduced extraneous solutions - solutions which work in the squared equation but not in the original one. For example we're given the equation $\displaystyle x-4=0$.
If we square it we arrive at $\displaystyle x^2=16$ which has both 4 and -4 as "solutions" but we can see that -4 clearly doesn't work in the original equation.

You must also consider the domain of the original equation which is what Siron pointed out by saying that $\displaystyle 2m+15 \geq 0$.

In this example one of your solutions to the quadratic will be extraneous - you can find out which by subbing the values into the original equation

6. ## Re: Radical Equation help

@Drewm:
Have you solved this inequality now? ...

7. ## Re: Radical Equation help

There's NO inequality involved; just solve: m^2 - 2m - 15 = 0
Easy to factor: (m - 5)(m + 3) = 0
So m = 5 or m = -3; over and out!

8. ## Re: Radical Equation help

Originally Posted by Wilmer
There's NO inequality involved; just solve: m^2 - 2m - 15 = 0
Easy to factor: (m - 5)(m + 3) = 0
So m = 5 or m = -3; over and out!
I agree with $\displaystyle m=5$

Howevr, $\displaystyle m=-3$ is not a solution since if we put -3 into the original equation we get $\displaystyle \sqrt{2 \cdot -3 +15} = \sqrt{9} = 3 \neq -3$

9. ## Re: Radical Equation help

Yes, offcourse, I was a bit confused, it's offcourse an equation.

10. ## Re: Radical Equation help

Originally Posted by e^(i*pi)
Howevr, $\displaystyle m=-3$ is not a solution since if we put -3 into the original equation we get $\displaystyle \sqrt{2 \cdot -3 +15} = \sqrt{9} = 3 \neq -3$
Original is really +/-SQRT(2m + 15) = +/-m ?

Anyway, IF the problem was to solve m^2 - 2m - 15 = 0 (no mention of SQRT...),
then we wouldn't care, right?

11. ## Re: Radical Equation help

Originally Posted by Wilmer
Original is really +/-SQRT(2m + 15) = +/-m ?
I don't think so, we have to reject $\displaystyle m=-3$. Because we may only square both sides if they're both positive and that's why we said $\displaystyle m>0$. And also if the equation would be:
$\displaystyle -\sqrt{2m+15}=-m$, we would wright this as:
$\displaystyle m=\sqrt{2m+15}$
Originally Posted by Wilmer
Anyway, IF the problem was to solve m^2 - 2m - 15 = 0 (no mention of SQRT...),
then we wouldn't care, right?
Right!