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# Why do square roots have two solutions and cube roots only have one solution?

Updated: 4/28/2022

Wiki User

8y ago

Square roots only have two solutions if the number is positive, if the number is negative it has no solutions.

Actually ALL numbers (including negative numbers) have two square roots and three cube roots - just not all of them are real numbers; some are complex numbers.

What is a complex number?

A complex number is a number which is the sum of two parts: a real part and an imaginary part, ie are of the form (a + bi):

a is the real part;

bi is the imaginary part which is a real number (b) multiplied by the square root of -1 which is an imaginary value. To avoid having to write √-1 all the time, the little 'i' is used instead, ie i² = -1.

Now the square roots of negative numbers can be found:

eg √-4 = √(4 × -1) = √4 √-1 = 2i → the square root of -4 is ±2i.

³√8 = 2 is the real root, but there are two further complex roots: (-1 + √3 i) and (-1 - √3 i). I'll cube the first to show it does indeed equal 8 (remember that i² = -1)

(-1 + √3 i)³ = (-1 + √3 i)(-1 + √3 i)(-1 + √3 i)

= (-1 + √3 i)((-1)×(-1) + -1 × √3 i + -1 × √3 i + (√3 i)×(√3 i))

= (-1 + √3 i)(1 - 2 (√3 i) + 3i²)

= (-1 + √3 i)(1 - 2 (√3 i) + 3 × -1)

= (-1 + √3 i)(1 - 2 (√3 i) - 3)

= (-1 + √3 i)(-2 - 2 (√3 i))

= (-1 + √3 i)(-1 - √3 i) × 2

= ((-1)² - (√3 i)²) × 2

= (1 - (-3)) × 2

= (1 + 3) × 2

= 4 × 2

= 8

Similar cubing of (-1 - √3 i) equals 8.

Wiki User

8y ago

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Q: Why do square roots have two solutions and cube roots only have one solution?
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