Please Solve This Urgently!

[Lwandile Hustler Thebe]

Guys i'm stuck. Solve this for me please, 3x=(x^1/2)+2

 

[RoaringInferno]

Assume x^1/2 as "a"
Thus 3x= x^1/2 + 2 becomes 3a^2 = a + 2
3a^2 - a -2 = 0
3a^2 - 3a + 2a - 2 = 0
(a - 1)(3a + 2) = 0
Therefore a = 1 OR a = -2/3
Since a = x^1/2
Therefore...
x = a^2
Thus...
x = (1)^2= 1 when a = 1
x = (-2/3)^2 = 4/9 when a = -2/3
Hope that helped.

 

[IGCSE O/L student]

Guys i'm stuck. Solve this for me please, 3x=(x^1/2)+2

3x-2=x/2
2(3x-2)=x
6x-4=x
5x=4
x=4/5

 

[RoaringInferno]

Guys i'm stuck. Solve this for me please, 3x=(x^1/2)+2

Is it x upon 2 or x (power 1/2)?

 

[Lwandile Hustler Thebe]

Don't worry, u already solved it. Thanks Again

 

[Lwandile Hustler Thebe]

You the best! Thank u so much.

 

[Lwandile Hustler Thebe]

You the best! Thank u so much.

 

[RoaringInferno]

Glad I could help ^_^

 

[Snowysangel]

Thus 3x= x^1/2 + 2 becomes 3a^2 = a + 2
[/end quotes)

Could someone please tell me how?? :/ this is log right? Ugh i suck at it

 

[IGCSE O/L student]

i'm not sure, but this might be right:
2a^2 = 2
2log2a = log 2
a= (log 2)/(2 log 2)

 

[scouserlfc]

Yeah this does come under logs but u actually dont require to use them here u can use the substitution method here that u must have learnt

 

[RoaringInferno]

Does that mean I'm correct?

 

[Khadija Sultana]

yah...u r correct!!!!!!!!!!!

 

[RoaringInferno]

;D*legendary feeling*

 

[Snowysangel]

Could someone explain this sum to me?


Why do you need the midpoints of PQ to find the equation of its perpendicular bisector?
Thank you in advance

 

[Dashing Cancerian]

Could someone explain this sum to me?
View attachment 13811

Why do you need the midpoints of PQ to find the equation of its perpendicular bisector?
Thank you in advance

cox the midpoint will be where the perpendicular bisector is. The perpendicular bisector bisects a line at its mid equally.

 

[anishh]

Guys i'm stuck. Solve this for me please, 3x=(x^1/2)+2

3x=(x^1/2)+2:

let y=x^1/2
so 3y^2=y+2
3y^2-y-2=o
(3y+2)(y-1)=0
y=-2/3 or y=1
x^1/2=-2/3 or x^1/2=1
x=(-2/3)^2 or x=1^2
x=4/9 or x=1