Radical (Math)
See how to solve a radical expression/equation/inequality.
32 examples and their solutions.
Square Root
Example
√9
Solution √9
= √32
= 3
= √32
= 3
Close
Example
-√25
Solution - √25
= - √52
= -5
= - √52
= -5
Close
Cube Root
Example
3√27
Solution 3√27
= 3√33
= 3
= 3√33
= 3
Close
Example
3√-8
Solution 3√-8
= 3√(-2)3
= -2
= 3√(-2)3
= -2
Close
Example
-3√-1000
Solution - 3√-1000
= - 3√(-10)3
= - (-10)
= 10
= - 3√(-10)3
= - (-10)
= 10
Close
Simplifying a Radical
Example
√4x2(x > 0)
Solution √4x2
= √22⋅x2
= 2 x
= √22⋅x2
= 2 x
Close
Example
√5a2bc6(a, b, c > 0)
Solution √5a2bc6
= √5⋅a2⋅b⋅(c3)2
= a c3 √5b
= √5⋅a2⋅b⋅(c3)2
= a c3 √5b
Close
Example
√12x9(x > 0)
Solution √12x9
= √22⋅3⋅(x4)2⋅x- [1] [2]
= 2 x4 √3x
= √22⋅3⋅(x4)2⋅x- [1] [2]
= 2 x4 √3x
[1]
12 = 22⋅3
Factoring (Math)
Factoring (Math)
[2]
x9 = x4⋅2 + 1 = x4⋅2⋅x = (x4)2⋅x
Exponent Rules
Exponent Rules
Close
Adding and Subtracting Radicals
Example
3√2 + 2√5 + 5√2 - 8√5
Solution 3√2 + 2√5 + 5√2 - 8√5
= 8√2 - 6√5
= 8√2 - 6√5
3 + 5 = 8
2 - 8 = -6
2 - 8 = -6
Close
Example
√50 + √18 - √2
Solution √50 + √18 - √2
= √2⋅52 + √2⋅32 - √2
= 5√2 + 3√2 - √2
= 7√2 - [1]
= √2⋅52 + √2⋅32 - √2
= 5√2 + 3√2 - √2
= 7√2 - [1]
[1]
5 + 3 - 1 = 7
Close
Multiplying Radicals
Example
√6x × √3x3y(x, y > 0)
Solution √6x × √3x3y
= √6x × 3x3y
= √2⋅3⋅x⋅3⋅x3⋅y
= √2⋅32⋅x4⋅y
= √2⋅32⋅(x2)2⋅y
= 3 x2 √2y
= √6x × 3x3y
= √2⋅3⋅x⋅3⋅x3⋅y
= √2⋅32⋅x4⋅y
= √2⋅32⋅(x2)2⋅y
= 3 x2 √2y
Close
Example
(√3 + √2)(√6 - 2)
Solution (√3 + √2)(√6 - 2)
= √3⋅√6 - 2√3 + √2⋅√6 - 2√2
= √3⋅√2⋅√3 - 2√3 + √2⋅√2⋅√3 - 2√2- [1]
= 3√2 - 2√3 + 2√3 - 2√2
= √2
= √3⋅√6 - 2√3 + √2⋅√6 - 2√2
= √3⋅√2⋅√3 - 2√3 + √2⋅√2⋅√3 - 2√2- [1]
= 3√2 - 2√3 + 2√3 - 2√2
= √2
[1]
√6 = √2⋅√3
Close
Dividing Radicals
Example
√10xy3√2y(x, y > 0)
Solution √10xy3√2y
= √10xy32y
= √5xy2
= y√5x
= √10xy32y
= √5xy2
= y√5x
Close
Rationalizing the Denominator
Example
√7y√x
Solution √7y√x⋅√x√x
= √7xyx
= √7xyx
Close
Example
√0.2
Solution √0.2
= √210
= √15
= 1√5
= 1√5⋅√5√5
= √55
= √210
= √15
= 1√5
= 1√5⋅√5√5
= √55
Close
Example
14 + √3
Solution 14 + √3⋅4 - √34 - √3- [1]
= 4 - √316 - 3- [2]
= 4 - √313
= 4 - √316 - 3- [2]
= 4 - √313
[1]
Denominator: a + √b
→ ×[a - √b]/[a - √b]
→ ×[a - √b]/[a - √b]
[2]
Close
Example
√25 - √6
Solution √25 - √6⋅5 + √65 + √6- [1]
= 5√2 + √2⋅√625 - 6
= 5√2 + √2⋅√2⋅√319
= 5√2 + 2√319
= 5√2 + √2⋅√625 - 6
= 5√2 + √2⋅√2⋅√319
= 5√2 + 2√319
[1]
Denominator: a - √b
→ ×[a + √b]/[a + √b]
→ ×[a + √b]/[a + √b]
Close
nth Root
Example
4√81
Solution 4√81
= 4√34
= 3
= 4√34
= 3
Close
Example
5√-32
Solution 5√-32
= 5√(-2)5
= -2
= 5√(-2)5
= -2
Close
Example
√x2
Solution √x2
= |x|
= |x|
√x2
x can be (-).
If x is (-),
the given, √x2, is (+).
But the result x is (-).
Then, to make the signs the same,
write the absolute value sign
to the result x.
x can be (-).
If x is (-),
the given, √x2, is (+).
But the result x is (-).
Then, to make the signs the same,
write the absolute value sign
to the result x.
Close
Example
4√16x12y8
Solution 4√16x12y8
= 4√24⋅(x3)4⋅(y2)4
= 2 |x3| y2 - [1]
= 4√24⋅(x3)4⋅(y2)4
= 2 |x3| y2 - [1]
[1]
4√16x12y8
x can be (-).
If x is (-),
the given, 4√16x12y8, is (+).
But 2x3y2 is (-).
Then, to make the signs the same,
x3 → |x3|.
Absolute Value
x can be (-).
If x is (-),
the given, 4√16x12y8, is (+).
But 2x3y2 is (-).
Then, to make the signs the same,
x3 → |x3|.
Absolute Value
Close
Rational Exponent
Formula
n√am = amn
Example
3√16
Solution 3√16
= 3√24
= 243
= 3√24
= 243
Close
Example
a79 = ?
Solution a79
= 9√a7
= 9√a7
Close
Example
452
Solution 452
= (22)52
= 22⋅52
= 25
= 32
= (22)52
= 22⋅52
= 25
= 32
Close
Example
Example
3√x2⋅6√x√x
Solution 3√x2⋅6√x√x
= x23⋅ x16x12
= x23 + 16 - 12
= x46 + 16 - 36
= x26
= x13
= 3√x
= x23⋅ x16x12
= x23 + 16 - 12
= x46 + 16 - 36
= x26
= x13
= 3√x
Close
Radical Equation
Example
√x + 6 = x
Solution √x + 6 = x
x + 6 ≥ 0- [1]
x ≥ -6
x ≥ 0- [2]
→ x ≥ 0- [3]
x + 6 = x2- [4]
-x2 + x + 6 = 0
x2 - x - 6 = 0
(x + 2)(x - 3) = 0- [5]
1) x + 2 = 0
x = -2
2) x - 3 = 0
x = 3- [6]
x = 3
x + 6 ≥ 0- [1]
x ≥ -6
x ≥ 0- [2]
→ x ≥ 0- [3]
x + 6 = x2- [4]
-x2 + x + 6 = 0
x2 - x - 6 = 0
(x + 2)(x - 3) = 0- [5]
1) x + 2 = 0
x = -2
2) x - 3 = 0
x = 3- [6]
x = 3
[1]
x + 6 is in the square root sign.
So x + 6 ≥ 0.
So x + 6 ≥ 0.
[2]
√x + 6 = x
The left side is (+).
So the right side, x, is (+).
The left side is (+).
So the right side, x, is (+).
[3]
Graph x ≥ -6, x ≥ 0 on a number line.
Color the intersecting region.
The intersecting region is x ≥ 0.
Color the intersecting region.
The intersecting region is x ≥ 0.
[4]
√x + 6 = x
Square both sides.
Square both sides.
[7]
Draw the condition x ≥ 0 on a number line.
Then see if x = -2, 3 are in the colored region.
x = -2 is not in the colored region.
So x = -2 is not the root.
x = 3 is in the colored region.
So x = 3 is the root.
Then see if x = -2, 3 are in the colored region.
x = -2 is not in the colored region.
So x = -2 is not the root.
x = 3 is in the colored region.
So x = 3 is the root.
Close
Radical Inequality
Example
√2x - 3 < 5
Solution √2x - 3 < 5
2x - 3 ≥ 0- [1]
2x ≥ 3
x ≤ 32
2x - 3 < 25- [2]
2x < 28
x < 14
32 ≤ x < 14
2x - 3 ≥ 0- [1]
2x ≥ 3
x ≤ 32
2x - 3 < 25- [2]
2x < 28
x < 14
32 ≤ x < 14
[1]
2x - 3 is in the square root sign.
So 2x - 3 ≥ 0.
So 2x - 3 ≥ 0.
[2]
√2x - 3 < 5
Square both sides.
Square both sides.
[3]
Draw the condition, x ≥ 3/2,
and the inequality you found, x < 14,
on a number line.
Then color the intersecting region.
and the inequality you found, x < 14,
on a number line.
Then color the intersecting region.
Close
Example
√x > √x - 2 + 1
Solution √x > √x - 2 + 1
x ≥ 0
x - 2 ≥ 0
x ≥ 2
→ x ≥ 2- [1]
x > (x - 2) + 2⋅√x - 2 ⋅1 + 1- [2]
0 > 2√x - 2 - 1
1 > 2√x - 2
2√x - 2 < 1
√x - 2 < 12
x - 2 < 14
x < 14 + 2
x < 14 + 84
x < 94
2 ≤ x < 94
x ≥ 0
x - 2 ≥ 0
x ≥ 2
→ x ≥ 2- [1]
x > (x - 2) + 2⋅√x - 2 ⋅1 + 1- [2]
0 > 2√x - 2 - 1
1 > 2√x - 2
2√x - 2 < 1
√x - 2 < 12
x - 2 < 14
x < 14 + 2
x < 14 + 84
x < 94
2 ≤ x < 94
[1]
Graph x ≥ 0, x ≥ 2 on a number line.
Color the intersecting region.
The intersecting region is x ≥ 0.
Color the intersecting region.
The intersecting region is x ≥ 0.
[2]
√x > √x - 2 + 1
Square both sides.
(√x - 2 + 1)2
= (√x - 2)2 + 2⋅√x - 2 ⋅1 + 12
= (x - 2) + 2⋅√x - 2 ⋅1 + 1
Factoring (Math)
Square both sides.
(√x - 2 + 1)2
= (√x - 2)2 + 2⋅√x - 2 ⋅1 + 12
= (x - 2) + 2⋅√x - 2 ⋅1 + 1
Factoring (Math)
[3]
Draw the condition, x ≥ 2,
and the inequality you found, x < 9/4,
on a number line.
Then color the intersecting region.
and the inequality you found, x < 9/4,
on a number line.
Then color the intersecting region.
Close
Square Root Function: Graph
Graph: y = √x
Graph of the Other Square Root Functions
Formula
y = √a(x - h) + k
Example
y = √2x - 6 + 1
Domain?
Solution Domain?
y = √2x - 6 + 1
y = √2(x - 3) + 1
x ≥ 3- [2]
y = √2(x - 3) + 1
x ≥ 3- [2]
[1]
Roughly graph y = √2(x - 3) + 1
on a coordinate plane.
on a coordinate plane.
[2]
The graph covers x ≥ 3.
So the domain is x ≥ 3.
So the domain is x ≥ 3.
Close
Example
y = √-x + 4
Domain?
Solution Domain?
y = √-x + 4
y = √-(x - 4)
x ≤ 4
y = √-(x - 4)
x ≤ 4
Close
Example
y = -√x + 2
Domain?
Solution Domain?
y = -√x + 2
x ≥ 0
x ≥ 0
Close