Solve Algebra Question: radical{n+4} + radical{n-1}=5

[redshadow]

Greetings guys!

I'm having a bit of trouble with an equation right now, I know the answer is 5, however, the steps I have to take to get there are pretty uncertain to me. Any help would be awesome.

radical{n+4} + radical{n-1}=5.


I need to know where to start. Impulse tells me to square the whole thing...but that doesn't seem to be working for me.

 

[Moo Of Doom]

Impulse takes me to multiply both sides by $$\left(\sqrt{n+4} - \sqrt{n-1}\right)$$, the "conjugate" of the left side. This is very common practice when dealing with square roots. Remember the difference of squares formula? That's the idea here.

 

[AKG]

Right now you have two radicals on the left side. Square both sides, and you'll only have one radical on the left side. Isolate it, and square both sides again. Your equation will now be free of radicals.

 

[AlphaNumeric]

Impulse takes me to multiply both sides by $$\left(\sqrt{n+4} - \sqrt{n-1}\right)$$, the "conjugate" of the left side. This is very common practice when dealing with square roots. Remember the difference of squares formula? That's the idea here.

The problem is that doesn't simplify it down, you still end up with the sum of two square roots in your expression. AKG's method is the way to go.

 

[HallsofIvy]

Actually, I would have done it slightly different from AKG's method:
First remove one radical to the right side:
$$\sqrt{n+4}= 5- \sqrt{n-1}$$
Now square both sides:
$$n+ 4= 25- 10\sqrt{n-1}+ n-1$$
Now the "n"s happen to cancel. Get the square root by itself on the left:
$$10\sqrt{n-1}= 20$$
or
$$\sqrt{n-1}= 2$$
Squaring that gives a simple equation for n.

(If the coefficients of "n" in the orginal equation had not been the same, the "n"s would not cancel after the first squaring. Then the final equation would be quadratic rather than linear.)

 

[VietDao29]

The problem is that doesn't simplify it down, you still end up with the sum of two square roots in your expression. AKG's method is the way to go.

HOI's, and AKG's approach are the common ones, however, you can also try out what Moo of Doom suggested.
Let $$I = \sqrt{n + 4} \quad \mbox{and} \quad J = \sqrt{n - 1}$$ (note that: I, J >= 0)
We have:

$$\sqrt{n + 4} + \sqrt{n - 1} = I + J = 5$$ (as the problem stated)

Now, if you multiply both sides with the conjugate of $$\sqrt{n + 4} + \sqrt{n - 1}$$, we'll obtain another equation:

$$(\sqrt{n + 4} + \sqrt{n - 1}) (\sqrt{n + 4} - \sqrt{n - 1}) = 5 ( \sqrt{n + 4} - \sqrt{n - 1} )$$

$$\Leftrightarrow 5 = 5 ( \sqrt{n + 4} - \sqrt{n - 1} )$$

$$\Leftrightarrow 1 = 1 ( \sqrt{n + 4} - \sqrt{n - 1} ) = I - J$$

Now, you have the system of equations:
$$\left\{ \begin{array}{l} I + J = 5 \\ I - J = 1 \end{array} \right.$$
Solve the system above for I, and J, then from there, you can find n. :)

 

[minase]

Viet how would you use this method if you have 3 radicals.

 

[VietDao29]

Viet how would you use this method if you have 3 radicals.

Like I said, it's not a common way. You can use it, but HOI's or AKG's approach may be more familiar. In a 3 radicals problem, you'll have 3 equations, and 3 unknowns (i.e 3 radicals).

 

[minase]

But how can you find the tree equations is there any way to find the conjugates of 3 radicals.