Question

solve for c
1/2(sqrt c-2) = sqrt 5/5

Answer 1

need algebra help

Answer 2

\[1/(2(\sqrt{c-2))}=\sqrt{5}/5\]

Answer 3

\[\frac{1}{2\sqrt{c-2}}=\frac{\sqrt{5}}{5}\]Is that correct?

Have you tried cross-multiplication and solving for \(c\)?

Answer 4

@whpalmer4 When you finish here, help me with my question?

Answer 5

ok ill try

Answer 6

that is correct

Answer 7

equation

Answer 8

not getting right answer

Answer 9

cross multiply =
\[2(\sqrt{5(c-2)}/5\]

Answer 10

No.

\[\frac{a}{b} = \frac{c}{d}\]cross-multiply:
\[a*d = b*c\]

I don't see an = anywhere in your work

Answer 11

\[5=2\sqrt{5(c-2)}\]
\[\sqrt{5}=2[5(c-2)]\]

Answer 12

does that look better so far?

Answer 13

sqrt5=10-2c-4
not sure what to do

Answer 14

6sgrt5/2=c=3sqrt5

Answer 15

which is incorrect

Answer 16

\[\frac{1}{2\sqrt{c-2}}=\frac{\sqrt{5}}{5}\]\[5*1 = \sqrt{5}*2\sqrt{c-2}\]\[5 = 2\sqrt{5}\sqrt{c-2}\]

Answer 17

At this point, why don't you square both sides?

Answer 18

i did I though?

sqrt5=2[5(c-2)]

Answer 19

\[5*5 = 2*2*\sqrt{5}*\sqrt{5}*\sqrt{c-2}*\sqrt{c-2}\]\[25 = 4*5*(c-2)\]

Answer 20

sqrt 5=10-2c-4

Answer 21

oh

Answer 22

\(a\) squared = \(a^2 = a*a\)

Answer 23

25=20(c-2)
25=20c-40
65=20c
13/4=c

Answer 24

\[(\sqrt{a})^2 = a\](assuming \(a\ge0\))

Answer 25

thanks

Answer 26

was t

Answer 27

was taking sqrt root instead of squaring

Answer 28

how do I know when to use correct method?

Answer 29

Yes, now you should check your answer in the original equation. Whenever you square both sides to eliminate a radical, you need to test any solutions to make sure they are not extraneous (read: don't work in the original equation).

Answer 30

original q is
find c mean value theorem f(x)=(sqrt x-2) [2,7]

Answer 31

so 13/4 = 3 1/4 is in interval is that what you mean by checking in original question

Answer 32

No, I mean you plug c = 13/4 into the original equation, and make sure the result is true.

It isn't an issue in this problem, because you had a radical on both sides, but if you solve something like \[2\sqrt{x-1} = x-3\] you get two solutions, but only one of them works as a solution to the original equation.

Answer 33

ok thanks

Answer 34

See this video for examples: https://www.khanacademy.org/math/algebra/exponent-equations/radical_equations/v/extraneous-solutions-to-radical-equations

Answer 35

@whpalmer4 Waiting whenever your ready :-)