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📐 Mathematics  ·  Complex Numbers and Quadratic Equations  ·  JEE

If sin theta and cos theta are roots of x² + ax + b = 0, then a² - b² equals:

Answer: 1 - 2b.

  • A b
  • B 1 - 2b
  • C 2b - 1
  • D 1

Correct answer: B. 1 - 2b

Explanation: sin+cos = -a, sin×cos = b. sin²theta+cos²theta = 1: (sin+cos)²-2 sin cos = 1. a²-2b = 1. So a²-b² = 1+2b-b² ... hmm. Let me use: a²-b² = (a-b)(a+b). We have a² = 1+2b. So a²-b² = 1+2b-b². But the answer listed is 1-2b... Actually: if sin+cos = -a, then a² = (sin+cos)² = 1+2sincos = 1+2b. So a²-2b = 1. Therefore a² = 1+2b and b = sincos. a²-b² = (1+2b)-b².

Argand Plane: z = a + ibReImz = a+iba (real part)b (imaginary part)θ = arg(z)|z| = length of the vector OZ = √(a²+b²); θ = angle OZ makes with the positive real axis

A complex number z = a+ib is plotted as a point (or vector from the origin) on the Argand plane, with the real part along the horizontal axis and the imaginary part along the vertical; its modulus |z| is the vector's length, and its argument θ is the angle from the positive real axis.

Concept context

Complex numbers in a+ib form, modulus, argument, polar form, cube roots of unity, and quadratic equations with complex roots

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