Algebra Question

1. Feb 16, 2005

DB

When

$$\sqrt{16 + (y-6)^2}=\sqrt{9+(y-5)^2}$$

is simplified to:

$$16+(y-6)^2=9+(y-5)^2$$

How come the 16 and 9 arent square rooted?

Thanks

2. Feb 16, 2005

Muzza

Do you agree that $\sqrt{a} = \sqrt{b}$ implies a = b?

If so, what happens when you set a = 16 + (y - 6)^2 and b = 9 + (y - 5)^2?

3. Feb 16, 2005

DB

ahhh, o ya its because they are equal, i got mixed up thanks.
But $$\sqrt{16 + (y-6)^2}$$
doesn't simplify to $$16+(y-6)^2$$ right?

4. Feb 16, 2005

Muzza

Correct. It would have if the expression was $\sqrt{ (16 + (y - 6)^2)^2 }$ though ;)

5. Feb 16, 2005

DB

ya thanks muzza

6. Feb 16, 2005

HallsofIvy

Staff Emeritus
If you square both sides of the equation,
$$\sqrt{16- (y-6)^2}= \sqrt{9- (y-5)^2}$$
you eliminate the two square roots and you get
16- (y-6)2= 9- (y-5)2.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook