Complex Numbers and Quadratic Equations Questions and Answers

Answer: C

The correct match is (P) → (1), (Q) → (2), (R) → (4), (S) → (5).

The roots of \(x^{2} + x + 1 = 0\) are the cube roots of unity \(\omega, \omega^{2}\), with \(\omega^{3} = 1\) and \(1 + \omega + \omega^{2} = 0\). So \(\omega + 1 = -\omega^{2}\) and \(\omega^{2} + 1 = -\omega\).

(P) \((\omega + 1)^{2026} = \omega^{4052} = \omega^{2}\) (since \(4052 = 3 \times 1350 + 2\)), and \((\omega^{2} + 1)^{2026} = \omega^{2026} = \omega\). The roots are the reciprocals \(\omega\) and \(\omega^{2}\), so the equation is \(x^{2} + x + 1 = 0\).

(Q) With the odd power 2027 a minus sign survives: \((-\omega^{2})^{2027} = -\omega\) and \((-\omega)^{2027} = -\omega^{2}\). The roots are \(-\omega^{2}\) and \(-\omega\), with sum 1 and product 1, so the equation is \(x^{2} - x + 1 = 0\).

(R) The roots of \(x^{2} - x + 1 = 0\) are \(-\omega\) and \(-\omega^{2}\), so \(\gamma - 1 = \omega^{2}\) and \(\delta - 1 = \omega\). The sum is \(\frac{1}{\omega^{4052}} + \frac{1}{\omega^{2026}} = \omega + \omega^{2} = -1\).

(S) Put \(a = p + 1\), \(b = r + 1\). They satisfy \(y^{2} - y - 1 = 0\), so \(a + b = 1\) and \(ab = -1\):

\[ \begin{aligned} \frac{1}{a^{3}} + \frac{1}{b^{3}} &= \frac{(a + b)^{3} - 3ab(a + b)}{(ab)^{3}} \\ &= \frac{1 + 3}{-1} = -4 \end{aligned} \]

Only the third option has all four pairings. The first swaps R and S, the fourth also gets P and Q wrong, and the second gets P and Q wrong.

Answer: A

a must lie in (-1, 1).

The equation is a perfect square minus 1:

\[ \begin{aligned} (x - a)^{2} - 1 &= 0 \\ x &= a - 1, \; a + 1 \end{aligned} \]

Both roots lie between -2 and 2 when a - 1 is greater than -2 and a + 1 is less than 2, that is, a greater than -1 and less than 1. (-2, 2) is the interval for the roots, not for a, and (-3, 0) and (0, 3) let a root escape the interval.

Answer: C

The equation has 0 real roots.

Put y = x², which turns it into a quadratic in y:

\[ \begin{aligned} 2y^{2} + 5y + 3 &= 0 \\ (2y + 3)(y + 1) &= 0 \\ y &= -1 \text{ or } -\tfrac{3}{2} \end{aligned} \]

Both values are negative, and x² cannot be negative for real x, so there are no real roots. A quicker check: every term on the left is positive for real x, so the sum can never be 0.

Answer: B

λ = \(-\frac{1}{2}\).

Bring every term to one side and collect the x terms:

\[ \begin{aligned} x^{2} - x - 2\lambda x + \lambda &= 0 \\ x^{2} - (1 + 2\lambda)x + \lambda &= 0 \end{aligned} \]

The sum of the roots is 1 + 2λ. Setting it to zero gives λ = -1/2. +1/2 comes from a sign slip, and ±2 from inverting the fraction.

Answer: D

The condition is 2b² = 9ac.

Let the roots be α and 2α. Then:

\[ \begin{aligned} 3\alpha &= -\frac{b}{a}, \quad 2\alpha^{2} = \frac{c}{a} \\ \alpha &= -\frac{b}{3a} \\ 2\cdot\frac{b^{2}}{9a^{2}} &= \frac{c}{a} \\ 2b^{2} &= 9ac \end{aligned} \]

b² = 4ac is the condition for equal roots, and the other two are not homogeneous in a, b and c, so multiplying the equation through by a constant would change them, which cannot happen to a condition on its roots.