TO APPLY THE SQUARE ROOT PRINCIPLE
THE QUADRATIC EQUATION MUST BE WRITTEN
AS A
A = (a + b)2
TO SOLVE USING THE
SQUARE ROOT PRINCIPLE
Must have “perfect square” variable expression on
one side and constant on the other ;
note—the constant doesn’t have to be a perfect
square, just the variable expression must be
“perfect”
Examples:
x2 = 16
(x – 4)2 = 9
(2x – 1)2 = 5
Circle the “perfect squares”
X2 = 4
a2 = 8
16n2 + 2n = 4
(m + 3)2 + m = 10
36d2 = 1
(2k -1)2 = 3
APPLYING THE SQUARE ROOT PRINCIPLE
RATIONAL ROOTS
x2 = 16
𝒙𝟐 = ± 𝟏𝟔
x = ±4
APPLYING THE SQUARE ROOT PRINCIPLE
IRRATIONAL ROOTS
a2 = 8
𝒂𝟐 = ± 𝟖
𝒂 = ± 𝟖 𝒐𝒓 ± 𝟐 𝟐
Graph can be deceiving!!
Use the CALC functions to verify.
Answer is very close to 3 ( 9 ) as ( 8 )
should be! BUT IT IS NOT 3!!!
Because the graph is symmetrical to the y-axis, the other
irrational root is approximately -2.828427.
Verify with your calculator that ± 𝟖 𝒐𝒓 ± 𝟐 𝟐 is
approximately ±2.828427
APPLYING THE SQUARE ROOT PRINCIPLE
(m + 3)2 = 10
(𝒎 + 𝟑)𝟐 = ± 𝟏𝟎
m + 3 = ± 𝟏𝟎
m = -3 ± 𝟏𝟎
Verify with calculator
APPLY THE SQUARE ROOT PRINCIPLE
36d2 = 1
(2k -1)2 = 3
IN QUADRATIC EQUATIONS…
• Irrational roots ALWAYS come in
CONJUGATE PAIRS 𝒂 ± 𝒃
• If you have one irrational root, then
you will have another one!!
WHAT ARE THE SOLUTIONS TO THE EQUATION?
𝟑 𝟐
𝒙 =
𝟒
48
HINT: Isolate the x2, then apply
the square root principle.