Finding Quadratic Equation with Complex Root
A quadratic equation with real coefficients must have complex roots in conjugate pairs. If one root is given as \(z = -5 - i\), the other root must be its complex conjugate.
Identifying the Roots
- Given root: \(r_1 = -5 - i\)
- According to the Complex Conjugate Root Theorem, since the coefficients are real, the other root must be the conjugate of \(r_1\).
- Complex conjugate root: \(r_2 = \overline{-5 - i} = -5 + i\)
Calculating Sum and Product of Roots
For a quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots is \(-\frac{b}{a}\) and the product of the roots is \(\frac{c}{a}\). For a monic quadratic equation (\(a=1\)), the equation is \(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\).
- Sum of roots (\(r_1 + r_2\)): \((-5 - i) + (-5 + i) = -5 - 5 + (-i + i) = -10\)
- Product of roots (\(r_1 \times r_2\)):
\((-5 - i)(-5 + i)\)
This is in the form \((a-b)(a+b) = a^2 - b^2\). Here \(a = -5\) and \(b = i\).
\((-5)^2 - (i)^2 = 25 - (-1) = 25 + 1 = 26\)
Constructing the Quadratic Equation
Substitute the calculated sum and product into the standard form:
\(x^2 - (\text{Sum})x + (\text{Product}) = 0\) \(x^2 - (-10)x + 26 = 0\) \(x^2 + 10x + 26 = 0\)This equation matches Option 2.
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