- #1
velox_xox
- 34
- 0
More fun with radicals!
...As for the problem, I tend to get problems with this same form (but with slight differences such as two separate binomials underneath individual square roots) wrong. I'm not quite sure what I'm doing wrong, so I was hoping for some help. :D
Solve. If an equation has no real solution, say so.
[tex] \sqrt {y} + \sqrt {y + 5} = 5 [/tex]
Answer: [itex] 4 [/itex]
--
The first thing I have been taught to do is to isolate the radicals on one side. That's already done.
Next, square both sides to eliminate radicals. I have a feeling that something I'm doing here is incorrect (?)
[itex] (\sqrt {y} + \sqrt {y + 5})^2 = 5^2 [/itex]
[itex] y +y + 5 = 25 [/itex]
Solve for y.
[itex] 2y + 5 = 25 [/itex]
[itex] 2y = 20 [/itex]
[itex] y = 10 [/itex]
But, since you often have the possibility of extraneous roots; you have to test it out in the original formula. When I do that, I get 'no solution.'
[itex] \sqrt {10} + \sqrt {10 + 5} \stackrel{?}{=} 5 [/itex]
[itex] \sqrt {10} + \sqrt {15} \stackrel {?} {=} 5 [/itex]
Approximate values from here on:
[itex] 3.16 + 3.87 \stackrel {?} {\approx} 5 [/itex]
[itex] 7.03 \neq 5 [/itex]
So, how'd I mess it up this time? (*joking*) Thanks in advance for your help!
...As for the problem, I tend to get problems with this same form (but with slight differences such as two separate binomials underneath individual square roots) wrong. I'm not quite sure what I'm doing wrong, so I was hoping for some help. :D
Homework Statement
Solve. If an equation has no real solution, say so.
[tex] \sqrt {y} + \sqrt {y + 5} = 5 [/tex]
Answer: [itex] 4 [/itex]
Homework Equations
--
The Attempt at a Solution
The first thing I have been taught to do is to isolate the radicals on one side. That's already done.
Next, square both sides to eliminate radicals. I have a feeling that something I'm doing here is incorrect (?)
[itex] (\sqrt {y} + \sqrt {y + 5})^2 = 5^2 [/itex]
[itex] y +y + 5 = 25 [/itex]
Solve for y.
[itex] 2y + 5 = 25 [/itex]
[itex] 2y = 20 [/itex]
[itex] y = 10 [/itex]
But, since you often have the possibility of extraneous roots; you have to test it out in the original formula. When I do that, I get 'no solution.'
[itex] \sqrt {10} + \sqrt {10 + 5} \stackrel{?}{=} 5 [/itex]
[itex] \sqrt {10} + \sqrt {15} \stackrel {?} {=} 5 [/itex]
Approximate values from here on:
[itex] 3.16 + 3.87 \stackrel {?} {\approx} 5 [/itex]
[itex] 7.03 \neq 5 [/itex]
So, how'd I mess it up this time? (*joking*) Thanks in advance for your help!