You are Here: Home >< A-levels

How to solve this basic C1 problem?

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016
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

Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: September 17, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

How does exam reform affect you?

From GCSE to A level, it's all changing

Poll

All the essentials

Student life: what to expect

What it's really like going to uni

Essay expert

Learn to write like a pro with our ultimate essay guide.

Create a study plan

Get your head around what you need to do and when with the study planner tool.

Resources by subject

Everything from mind maps to class notes.

Study tips from A* students

Students who got top grades in their A-levels share their secrets