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Complex Numbers and Quadratic Equations - The square roots of a negative real number

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The imaginary unit ii is defined such that i2=−1i^2 = -1. This allows us to define the square root of negative numbers.

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For any positive real number aa, the square root of −a-a is expressed as −a=a⋅−1=ia\sqrt{-a} = \sqrt{a} \cdot \sqrt{-1} = i\sqrt{a}.

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Every negative real number has two square roots in the complex number system. For example, the square roots of −a-a (where a>0a > 0) are iai\sqrt{a} and −ia-i\sqrt{a}.

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The symbol −a\sqrt{-a} is generally used to denote the principal square root, which is iai\sqrt{a}.

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The property a⋅b=ab\sqrt{a} \cdot \sqrt{b} = \sqrt{ab} is valid ONLY if at least one of aa or bb is a non-negative real number. If both a<0a < 0 and b<0b < 0, then a⋅b=(i∣a∣)(i∣b∣)=i2∣ab∣=−ab\sqrt{a} \cdot \sqrt{b} = (i\sqrt{|a|})(i\sqrt{|b|}) = i^2\sqrt{|ab|} = -\sqrt{ab}.

📐Formulae

i=−1i = \sqrt{-1}

i2=−1i^2 = -1

−a=ia for any a>0\sqrt{-a} = i\sqrt{a} \text{ for any } a > 0

−a⋅−b=−ab where a,b>0\sqrt{-a} \cdot \sqrt{-b} = -\sqrt{ab} \text{ where } a, b > 0

💡Examples

Problem 1:

Find the square roots of −16-16.

Solution:

Let x2=−16x^2 = -16. Taking the square root on both sides: x=±−16x = \pm \sqrt{-16} x=±16⋅−1x = \pm \sqrt{16} \cdot \sqrt{-1} x=±4ix = \pm 4i

Explanation:

We identify a=16a = 16. The square roots are ±i16\pm i\sqrt{16}, which simplifies to ±4i\pm 4i.

Problem 2:

Evaluate the product −9×−4\sqrt{-9} \times \sqrt{-4}.

Solution:

First, express each square root in terms of ii: −9=3i\sqrt{-9} = 3i −4=2i\sqrt{-4} = 2i Now, multiply them: 3i×2i=6i23i \times 2i = 6i^2 Since i2=−1i^2 = -1: 6(−1)=−66(-1) = -6

Explanation:

Note that multiplying them directly as (−9)(−4)=36=6\sqrt{(-9)(-4)} = \sqrt{36} = 6 is incorrect because the identity ab=ab\sqrt{a}\sqrt{b} = \sqrt{ab} does not hold when both numbers are negative.

Problem 3:

Solve the quadratic equation x2+25=0x^2 + 25 = 0.

Solution:

x2=−25x^2 = -25 x=±−25x = \pm \sqrt{-25} x=±5ix = \pm 5i

Explanation:

Moving 2525 to the RHS makes it negative. Applying the definition of square roots of negative numbers gives the imaginary solutions.