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anonymous

  • one year ago

A quadratic equation with real coefficients and leading coefficient 1, has x = -bi as a root. Write the equation in general form.

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  1. mathmate
    • one year ago
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    A polynomial with real coefficients has complex roots in conjugate pairs. Hence if one root is -bi, the conjugate root is +bi. The quadratic is then f(x)=a(x+bi)(x-bi). If the leading coefficient (coefficient of the term of highest degree, namely x^2) is one, then a=1, giving f(x)=(x+bi)(x-bi), Expand the expression and write in general form to complete the solution.

  2. anonymous
    • one year ago
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    how will i write in in general form? :)

  3. mathmate
    • one year ago
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    See: https://www.mathsisfun.com/algebra/polynomials-general-form.html

  4. anonymous
    • one year ago
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    \[x ^{2}-bi^2 =0 ???

  5. mathmate
    • one year ago
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    Almost! But do not forget \(ab^2\) does not equal \((ab)^2=a^2b^2\), and that \(i^2=-1\) \(x ^{2}-(bi)^2 =0\) Expand the second term and simplify to get the general form.

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