two math problems

[cavanbasParticipant]

1. Given that 3-x1/2=(2x-7)1/2, what is x?
(i.e., “1/2” means square root.)

2. Given that x+y=11 and x1/2=y-5, solve for x and y.

1. Ans. x=4 or 64.
2. Ans. x=4, y=7; x=9, y=2.

[y82benjiParticipant]

3 – sqrt[x] = sqrt[2x-7]
square both sides
(3-sqrt[x])(3-sqrt[x]) = ([2x-7]^0.5)^2
multiply out on left, mulitply exponents on the right
3*3 – 3*sqrt[x] – 3*sqrt[x] + sqrt[x]*sqrt[x] = [2x-7]^1
simplify
9 – 6*sqrt[x] + x = 2x – 7
subtract 9 and x from both sides
-6*sqrt[x] = x – 16
square both sides
36x = x^2 – 32x + 256
subtract 36x from both sides
x^2 – 68x + 256
use quadratic
get x = 4 or 64

[y82benjiParticipant]

sqrt[x] = y – 5
square both sides
x = y^2 – 10y + 25
then
x + y = 11
subtract y from both sides
x = 11 – y
substitute “x = 11 – y” in x = y^2 – 10y + 25
11 – y = y^2 – 10y + 25
subtract 11 and add y to both sides
0 = y^2 – 9y + 14
use quadratic to solve and get
y = 2 or 7
go back to “x + y = 11”
x = 9 or 4

[jelloParticipant]

For the 1st question, the answer should be ONLY 4, not 64.
3-sqrt(64)=sqrt(2*64-7)
3-8=sqrt(121)
-5=11 which is incorrect.

Hence, the answer is only 4

[y82benjiParticipant]

No, 64 is still correct because the square root of 64 can be -8 or +8. If you take the square root of 64 to be -8 then you have 3 – -8 = 11 = sqrt(121)

[liebenatorParticipant]

I should not have read this thread. Now I have a headache.[]

[y82benjiParticipant]

Gotta love math. It’s just a language with more grammar than vocabulary!

[kaushalaspParticipant]

rightly said…all the best…

[auburn88Participant]

I have a headache too Liebenator. That makes the case for starting at the end of a thread and reading up.

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