Question

Solve the equation: √k+2-√k-3=1.

Answer 1

k=7 it says in the answer key, but when I worked the problem out, I got 7/2.

Answer 2

Is the square root over the entire (k+2) or just k. And over entire (k-3) or just k?
If it is over the entire thing you should use parenthesis to group items.

Answer 3

Over the entire k+2 and the entire k-3

Answer 4

That makes a huge difference.

Answer 5

Take one radical over to the other side. Square both sides. Then rearrange terms so that only the radical remains on one side. Square them again to get rid of the radical. Solve for k.

Answer 6

After you square it the first time, it looks like this: k+2=k-3+(2√k-3) +1 Right?

Answer 7

sqrt(k+2) - sqrt(k-3) = 1
sqrt(k+2) = 1 + sqrt(k-3)
square both sides
k+2 = 1 + k - 3 + 2sqrt(k-3) { (a+b)^2 = a^2 + b^2 + 2ab }
2 - 1 + 3 = 2sqrt(k-3)
4 = 2sqrt(k-3)
square them again and solve for k

Answer 8

4 = 2sqrt(k-3)
divide both sides by 2
2 = sqrt(k-3)
square both sides
4 = k - 3
k = 7

Answer 9

When you square the (2√k-3) how do you foil it out?

Answer 10

square 2 and then get rid of the radical

Answer 11

{ 2 * sqrt(k-3) }^2 = 2 * sqrt(k-3) * 2 * sqrt(k-3) = (2 * 2) * sqrt(k-3) * sqrt(k-3)
= 4 * (k-3) = 4k - 12

Answer 12

square root and square will cancel out. sqrt(5) * sqrt(5) = 5

Answer 13

How does the 12 get there? I was thinking it would look like: 16=4(k-3) after you square them the second time.

Answer 14

4(k-3) = 4 * k - 4 * 3

Answer 15

Yeah, I didn't notice that at first. Thank you so much for helping me!! I see where the 7 comes from now.

Answer 16

you are welcome.