If for z=α+iβ, |z+2|=z+4(1+i), then α +β and αβ are the roots of the equation
Step 1: Solve the given condition.
\[ |z + 2| = z + 4(1 + i). \] - Substitute \(z = \alpha + i\beta\): \[ |\alpha + i\beta + 2| = (\alpha + 4) + i(\beta + 4). \] This represents a complex number equation where \(z\) is expressed as \( \alpha + i\beta \).
Step 2: Solve for \(\alpha\) and \(\beta\).
- Expand the magnitudes: \[ \sqrt{(\alpha + 2)^2 + \beta^2} = \sqrt{(\alpha + 4)^2 + (\beta + 4)^2}. \] Now, equate the real and imaginary parts of both sides to solve for \( \alpha \) and \( \beta \): - After solving, we find: \[ \alpha = 1, \quad \beta = -4. \] Step 3: Find the quadratic equation.
- The roots are \( \alpha + \beta = -3 \) and \( \alpha \beta = -4 \). Thus, the quadratic equation is: \[ x^2 - (\alpha + \beta)x + \alpha\beta = x^2 + 7x + 12. \] Final Answer: The quadratic equation is \( x^2 + 7x + 12 = 0 \).
The obtuse angle between lines \(2y = x + 1\) and \(y = 3x + 2\) is:
What is the general solution of the equation \( \cot\theta + \tan\theta = 2 \)?
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to: