Digital SAT Math Practice Question #1192
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Remember the properties of polynomial roots when coefficients are real. If a complex number \( a+bi \) is a root, what other root must exist? How does the degree of the polynomial relate to the total number of roots?
Consider Vieta's formulas, specifically how the constant term \( D \) relates to the roots of a monic polynomial. How can \( P(0) \) help you find the missing root?
⚡ Hack the Test speed shortcut: 1. **Complex Conjugate Root Theorem:** Immediately identify \( 2+i \) as the second root because the polynomial has real coefficients. 2. **Vieta's Formulas for Constant Term:** For a monic polynomial \( P(x) = x^4 + Ax^3 + Bx^2 + Cx + D \), the constant term \( D \) is equal to \( P(0) \) and also the product of all roots. So, \( D = -24 \). Set up the product of the four roots: \( (2-i)(2+i)(\sqrt{3})r_4 = -24 \). Simplify the complex conjugate product: \( (4 - i^2)(\sqrt{3})r_4 = -24 \implies 5\sqrt{3}r_4 = -24 \). Solve for the fourth root: \( r_4 = \frac{-24}{5\sqrt{3}} = \frac{-24\sqrt{3}}{15} = \frac{-8\sqrt{3}}{5} \). 3. **Identify Real Roots and Sum:** The real roots are \( \sqrt{3} \) and \( \frac{-8\sqrt{3}}{5} \). Sum them: \( \sqrt{3} + \left(\frac{-8\sqrt{3}}{5}\right) = \frac{5\sqrt{3} - 8\sqrt{3}}{5} = \frac{-3\sqrt{3}}{5} \). This method avoids expanding the polynomial, which is time-consuming and error-prone, allowing for a quick and accurate solution.
The problem asks for the sum of the real roots of a polynomial \( P(x) = x^4 + Ax^3 + Bx^2 + Cx + D \) with real coefficients and a leading coefficient of 1. We are given two roots: \( 2-i \) and \( \sqrt{3} \), and \( P(0) = -24 \). **Step...
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