complex numbers and quadratic equations class 11 ncert solutions
What are complex numbers and quadratic equations?
In this article, we will discuss the complex numbers and quadratic equations and the nature of roots in detail. A complex number can be represented in the form of a+bi, which is the combination of both the real numbers and the imaginary numbers. Here, a and b are real numbers and i is the imaginary number.
Is a complex-valued answer a 'root' or'solution' for a quadratic equation?
Your complex-valued answer is still a valid "zero" or "root" or "solution" for that quadratic equation, because, if you plug the answer value back into the quadratic equation, you'll get zero after you simplify. (This is the definition of a "solution"; it is a value that solves the equation, making it true.)
Complex Numbers and Quadratic Equations
A complex number can be represented in the form of a+bi, which is the combination of both the real numbers and the imaginary numbers. Here, a and b are real numbers and i is the imaginary number. byjus.com
Solution of Complex Quadratic Equations
A quadratic equation is an equation, where atleast one term should be squared. The maximum degree of the equation must be two. For example, 5x2+3x+3 =0. In this case, the highest order of the equation is 2. So, the given equation is a quadratic equation. byjus.com
Solved Examples
Example 1: Find the roots of the quadratic equation Solution: Comparing the given equation with the general form of the equation We have, a= 1, b= -1, c= 1 D= Thus the equation have two complex roots. Thus the roots are Example 2: Solve Solution: Comparing the given equation with the general form of the equation We have, Here Discriminant (D) = byjus.com
Solving using the quadratic formula with complex solutions
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Since both the values of sin θand cos θ are negative and sinθand cosθare negative in III quadrant, Thus, the modulus and argument of the complex number |
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Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13 11 4 – 3i Solution: Let's consider z = 4 – 3i Then, = 4 + 3i and |
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Overall Hint: Take the complex number as z find the conjugate of z then apply the modulus of z Divide the conjugate with the square of modulus to give the |
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