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📐 Mathematics  ·  Complex Numbers and Quadratic Equations  ·  JEE

For what value of k does kx² + 6x + 1 = 0 have real roots?

Answer: k less than or equal to 9.

  • A k > 9, obtained by reversing the discriminant inequality
  • B k less than or equal to 9
  • C k < 0, requiring the leading coefficient to be negative
  • D k = 9 only, taken from setting the discriminant to zero

Correct answer: B. k less than or equal to 9

Explanation: D = 36 - 4k >= 0. 4k <= 36. k <= 9. Also k cannot be 0 (then not quadratic). k <= 9, k not 0.

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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