Transcript Document

Surd or Radical
Equations
To solve an equation with a surd
2x 1  3  4
+3
+3
2
2x 1  7
2 x  1  49
+1
+1
2 x  50
x  25
First isolate the surd
2
This means to get any terms not
under the square root on the other
side of the equal sign
Now square both sides
You must square the whole side
NOT each term. A square
"undoes"
or cancels
Now
solve
for x a square root
You MUST check this answer
225 1  3  4
Since you squared both sides of the
equation, negatives disappear. It is possible
to get an answer that
doesn't work when
It checks!
you plug it back in
44
Let's try another one:
2 x  1
1
3
2 x  1
1  0
-1
1 3
3
First isolate the surd
-1
 1
3
2 x  1  1
-1
-1
2 x  2
x  1
Remember that the 1/3
Now
is same
a 1/3thing
powersince
meansitthe
as a cube
root.
power
this
means the
same as a cube root so
cube both sides
Now solve for x
Let's check this answer
21 1 1  0
0  0 It checks!
3
One more to see extraneous solution:
surd is already
isolated
algebraically
but DOES
NOT
3axsolution
 1 that
x you3findThe
make a true statement when you substitute it back
2
2 Square both sides
into the equation.
You must square the whole side
NOT each term.
3x  1  x  3
This must be FOILed
3x  1  x  6 x  9 You MUST check
answers
2 you have a quadratic these
Since
equation
(has an x2
x

9
x

8

0
term) get everything on one side = 0 and see if


3
1

1

1

3
you can factor this
3
8


8

3
x  8x 1  0
Doesn't
work!
It
checks!
x  8, x  1
5252 Extraneous
2
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au