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# How to solve this basic C1 problem? Watch

1. a) x^1/2 + 2x^-1/2 = 3

b) x + x^1/2 = 12

I know the answers because I did trial and error in my head. If I am do to lines of working out, what steps lead to the answer?
2. (Original post by Baaah)
a) x^1/2 + 2x^-1/2 = 3

b) x + x^1/2 = 12

I know the answers because I did trial and error in my head. If I am do to lines of working out, what steps lead to the answer?
Let

Your part a turns into then multiply through by A and solve the quadratic in A. Then once you have turn it into and get x.

Same with part b. It would turn into
3. (Original post by Baaah)
a) x^1/2 + 2x^-1/2 = 3

b) x + x^1/2 = 12

I know the answers because I did trial and error in my head. If I am do to lines of working out, what steps lead to the answer?
a) To solve x^1/2 + 2x^-1/2 = 3

let Y = x^(1/2)

remember 2x^(-1/2) = 2/x^(1/2)

; Y + 2/Y = 3

*Y ; Y^2 + 2 = 3Y

; Y^2 - 3Y + 2 =0

; (Y - 1)(Y - 2) = 0

; Y = 1 , Y = 2

Substitute Y = 1 and Y = 2 into Y = x^(1/2)

; 1 = x^(1/2) ; 2 = x^(1/2)

; x = 1 ; x = 4

b) To solve x + x^1/2 = 12

let U = x^1/2

remember x = (x^1/2)^2

therefore ; U^2 + U = 12

; U^2 + U - 12 = 0

; (U + 4)(U - 3)

; U = -4 , U = 3

substitute U = -4 and U = 3 into U = x^1/2

; -4 = x^1/2 ; 3 = x^1/2

; x = no solution ; x = 9

Hope this helps
4. (Original post by Pixel Warrior)
a) To solve x^1/2 + 2x^-1/2 = 3

let Y = x^(1/2)

remember 2x^(-1/2) = 2/x^(1/2)

; Y + 2/Y = 3

*Y ; Y^2 + 2 = 3Y

; Y^2 - 3Y + 2 =0

; (Y - 1)(Y - 2) = 0

; Y = 1 , Y = 2

Substitute Y = 1 and Y = 2 into Y = x^(1/2)

; 1 = x^(1/2) ; 2 = x^(1/2)

; x = 1 ; x = 4

b) To solve x + x^1/2 = 12

let U = x^1/2

remember x = (x^1/2)^2

therefore ; U^2 + U = 12

; U^2 + U - 12 = 0

; (U + 4)(U - 3)

; U = -4 , U = 3

substitute U = -4 and U = 3 into U = x^1/2

; -4 = x^1/2 ; 3 = x^1/2

; x = no solution ; x = 9

Hope this helps
Damn, can I get you to do my homework for me as well???
5. (Original post by Pixel Warrior)
a) To solve x^1/2 + 2x^-1/2 = 3

let Y = x^(1/2)

remember 2x^(-1/2) = 2/x^(1/2)

; Y + 2/Y = 3

*Y ; Y^2 + 2 = 3Y

; Y^2 - 3Y + 2 =0

; (Y - 1)(Y - 2) = 0

; Y = 1 , Y = 2

Substitute Y = 1 and Y = 2 into Y = x^(1/2)

; 1 = x^(1/2) ; 2 = x^(1/2)

; x = 1 ; x = 4

b) To solve x + x^1/2 = 12

let U = x^1/2

remember x = (x^1/2)^2

therefore ; U^2 + U = 12

; U^2 + U - 12 = 0

; (U + 4)(U - 3)

; U = -4 , U = 3

substitute U = -4 and U = 3 into U = x^1/2

; -4 = x^1/2 ; 3 = x^1/2

; x = no solution ; x = 9

Hope this helps
2x^(-1/2) = 2/x^(1/2)
I thought 2x^(-1/2) = 1/2x^(1/2) ?
6. nope
(Original post by Baaah)
2x^(-1/2) = 2/x^(1/2)
I thought 2x^(-1/2) = 1/2x^(1/2) ?
7. (Original post by Baaah)
2x^(-1/2) = 2/x^(1/2)
I thought 2x^(-1/2) = 1/2x^(1/2) ?
No that's wrong

by breaking it up you see that

2x^(-1/2) = 2 * x^(-1/2)

= 2 * 1/x^(1/2)

= 2/x^(1/2)
8. (Original post by Pixel Warrior)
No that's wrong

by breaking it up you see that

2x^(-1/2) = 2 * x^(-1/2)

= 2 * 1/x^(1/2)

= 2/x^(1/2)
Oh I see

I was confused because I thought a negative index meant I just have to make the whole base a reciprocal
9. (Original post by Baaah)
2x^(-1/2) = 2/x^(1/2)
I thought 2x^(-1/2) = 1/2x^(1/2) ?

Ahh, beaten to it. Never mind...
10. (Original post by K-Man_PhysCheM)

Ahh, beaten to it. Never mind...
Well thanks anyway

No wonder I kept getting it wrong...
11. (Original post by Pixel Warrior)
a) To solve x^1/2 + 2x^-1/2 = 3

let Y = x^(1/2)

remember 2x^(-1/2) = 2/x^(1/2)

; Y + 2/Y = 3

*Y ; Y^2 + 2 = 3Y

; Y^2 - 3Y + 2 =0

; (Y - 1)(Y - 2) = 0

; Y = 1 , Y = 2

Substitute Y = 1 and Y = 2 into Y = x^(1/2)

; 1 = x^(1/2) ; 2 = x^(1/2)

; x = 1 ; x = 4

b) To solve x + x^1/2 = 12

let U = x^1/2

remember x = (x^1/2)^2

therefore ; U^2 + U = 12

; U^2 + U - 12 = 0

; (U + 4)(U - 3)

; U = -4 , U = 3

substitute U = -4 and U = 3 into U = x^1/2

; -4 = x^1/2 ; 3 = x^1/2

; x = no solution ; x = 9

Hope this helps
Sorry to bother you again

-4 = x^1/2
Wouldn't that mean x has a solution of 16 because (-4)^2 = 16? Why is there no solution :?
12. (Original post by Baaah)
Sorry to bother you again

-4 = x^1/2
Wouldn't that mean x has a solution of 16 because (-4)^2 = 16? Why is there no solution :?
I guess that since sqrt(16) gives +/- 4 you are correct, but you should ask a teacher or something as its been a while since i've done this.

And no probs, happy to help
13. (Original post by Baaah)
Sorry to bother you again

-4 = x^1/2
Wouldn't that mean x has a solution of 16 because (-4)^2 = 16? Why is there no solution :?
No because is always greater than or equal to 0.

Besides, when you plug it back through the equation.
14. (Original post by Pixel Warrior)
I guess that since sqrt(16) gives +/- 4 you are correct, but you should ask a teacher or something as its been a while since i've done this.

And no probs, happy to help

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