• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Square Root Graph Question (1 Viewer)

Ambility

Active Member
Joined
Dec 22, 2014
Messages
336
Gender
Male
HSC
2016
Why is the graph only above the x-axis? Shouldn't this have a mirror reflection below the x-axis?

For example, the square root of 25 is 5, but also -5. Why is this not in the graph?
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
False.

The square root of 25 is 5.
√25 = 5

But the equation x^2 = 25 has two solutions
|x| = 5
x = ±5

Note that -√25 = -5. When we take the square root OPERATOR we only consider the principal root.
But the quadratic EQUATION will have two solutions.
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Additionally, consider the graph of x = y^2

This is a sideway parabola.

But the graph of y = √x is only the top branch of it.
 

Ambility

Active Member
Joined
Dec 22, 2014
Messages
336
Gender
Male
HSC
2016

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Last edited:

Nailgun

Cole World
Joined
Jun 14, 2014
Messages
2,193
Gender
Male
HSC
2016
From your calculator webpage lol

Any nonnegative real number x has a unique nonnegative square root r; this is called the principal square root .......... For example, the principal square root of 9 is sqrt(9) = +3, while the other square root of 9 is -sqrt(9) = -3. In common usage, unless otherwise specified, "the" square root is generally taken to mean the principal square root."[1].
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Why is the graph only above the x-axis? Shouldn't this have a mirror reflection below the x-axis?

For example, the square root of 25 is 5, but also -5. Why is this not in the graph?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top