Radical (Math)
See how to solve a radical expression/equation/inequality.
32 examples and their solutions.
Square Root
Example
√9
Solution √9
= √3^{2}
= 3
= √3^{2}
= 3
Close
Example
-√25
Solution - √25
= - √5^{2}
= -5
= - √5^{2}
= -5
Close
Cube Root
Example
^{3}√27
Solution ^{3}√27
= ^{3}√3^{3}
= 3
= ^{3}√3^{3}
= 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
√4x^{2}(x > 0)
Solution √4x^{2}
= √2^{2}⋅x^{2}
= 2 x
= √2^{2}⋅x^{2}
= 2 x
Close
Example
√5a^{2}bc^{6}(a, b, c > 0)
Solution √5a^{2}bc^{6}
= √5⋅a^{2}⋅b⋅(c^{3})^{2}
= a c^{3} √5b
= √5⋅a^{2}⋅b⋅(c^{3})^{2}
= a c^{3} √5b
Close
Example
√12x^{9}(x > 0)
Solution √12x^{9}
= √2^{2}⋅3⋅(x^{4})^{2}⋅x- [1] [2]
= 2 x^{4} √3x
= √2^{2}⋅3⋅(x^{4})^{2}⋅x- [1] [2]
= 2 x^{4} √3x
[1]
12 = 2^{2}⋅3
Factoring (Math)
Factoring (Math)
[2]
x^{9} = x^{4⋅2 + 1} = x^{4⋅2}⋅x = (x^{4})^{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⋅5^{2} + √2⋅3^{2} - √2
= 5√2 + 3√2 - √2
= 7√2 - [1]
= √2⋅5^{2} + √2⋅3^{2} - √2
= 5√2 + 3√2 - √2
= 7√2 - [1]
[1]
5 + 3 - 1 = 7
Close
Multiplying Radicals
Example
√6x × √3x^{3}y(x, y > 0)
Solution √6x × √3x^{3}y
= √6x × 3x^{3}y
= √2⋅3⋅x⋅3⋅x^{3}⋅y
= √2⋅3^{2}⋅x^{4}⋅y
= √2⋅3^{2}⋅(x^{2})^{2}⋅y
= 3 x^{2} √2y
= √6x × 3x^{3}y
= √2⋅3⋅x⋅3⋅x^{3}⋅y
= √2⋅3^{2}⋅x^{4}⋅y
= √2⋅3^{2}⋅(x^{2})^{2}⋅y
= 3 x^{2} √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
√10xy^{3}√2y(x, y > 0)
Solution √10xy^{3}√2y
= √10xy^{3}2y
= √5xy^{2}
= y√5x
= √10xy^{3}2y
= √5xy^{2}
= 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}√3^{4}
= 3
= ^{4}√3^{4}
= 3
Close
Example
^{5}√-32
Solution ^{5}√-32
= ^{5}√(-2)^{5}
= -2
= ^{5}√(-2)^{5}
= -2
Close
Example
√x^{2}
Solution √x^{2}
= |x|
= |x|
√x^{2}
x can be (-).
If x is (-),
the given, √x^{2}, 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, √x^{2}, 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}√16x^{12}y^{8}
Solution ^{4}√16x^{12}y^{8}
= ^{4}√2^{4}⋅(x^{3})^{4}⋅(y^{2})^{4}
= 2 |x^{3}| y^{2} - [1]
= ^{4}√2^{4}⋅(x^{3})^{4}⋅(y^{2})^{4}
= 2 |x^{3}| y^{2} - [1]
[1]
^{4}√16x^{12}y^{8}
x can be (-).
If x is (-),
the given, ^{4}√16x^{12}y^{8}, is (+).
But 2x^{3}y^{2} is (-).
Then, to make the signs the same,
x^{3} → |x^{3}|.
Absolute Value
x can be (-).
If x is (-),
the given, ^{4}√16x^{12}y^{8}, is (+).
But 2x^{3}y^{2} is (-).
Then, to make the signs the same,
x^{3} → |x^{3}|.
Absolute Value
Close
Rational Exponent
Formula
^{n}√a^{m} = a^{mn}
Example
^{3}√16
Solution ^{3}√16
= ^{3}√2^{4}
= 2^{43}
= ^{3}√2^{4}
= 2^{43}
Close
Example
a^{79} = ?
Solution a^{79}
= ^{9}√a^{7}
= ^{9}√a^{7}
Close
Example
4^{52}
Solution 4^{52}
= (2^{2})^{52}
= 2^{2⋅52}
= 2^{5}
= 32
= (2^{2})^{52}
= 2^{2⋅52}
= 2^{5}
= 32
Close
Example
3^{-72}⋅ 3^{52}
Solution Example
^{3}√x^{2}⋅^{6}√x√x
Solution ^{3}√x^{2}⋅^{6}√x√x
= x^{23}⋅ x^{16}x^{12}
= x^{23 + 16 - 12}
= x^{46 + 16 - 36}
= x^{26}
= x^{13}
= ^{3}√x
= x^{23}⋅ x^{16}x^{12}
= x^{23 + 16 - 12}
= x^{46 + 16 - 36}
= x^{26}
= x^{13}
= ^{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 = x^{2}- [4]
-x^{2} + x + 6 = 0
x^{2} - x - 6 = 0
(x + 2)(x - 3) = 0- [5]
1) x + 2 = 0
x = -2
2) x - 3 = 0
x = 3- [6]
- [7]
x = 3
x + 6 ≥ 0- [1]
x ≥ -6
x ≥ 0- [2]
→ x ≥ 0- [3]
x + 6 = x^{2}- [4]
-x^{2} + x + 6 = 0
x^{2} - 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
- [3]
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
- [3]
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 + 1^{2}
= (x - 2) + 2⋅√x - 2 ⋅1 + 1
Factoring (Math)
Square both sides.
(√x - 2 + 1)^{2}
= (√x - 2)^{2} + 2⋅√x - 2 ⋅1 + 1^{2}
= (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
- [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