 # What is a quadratic equation in standard form with rational coefficients that has a root of 5 + 4i? arenceabigns 2021-01-06 Answered
What is a quadratic equation in standard form with rational coefficients that has a root of 5 + 4i?
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Complex zeros come in conjugate pairs so we know that 5-4i is a zero given that 5+4i is a zero.
If a is zero of a polynomial function, then x-a is one of its factor.
$y=a\left(x-\left(5+4i\right)\right)\left(x-\left(5-4i\right)\right)$
By expanding
$y=a\left[{x}^{2}-\left(5-4i\right)x-\left(5+4i\right)x+\left(5+4i\right)\left(5-4i\right)\right]$ $y=a\left[{x}^{2}-5x+4ix-5x-4ix+\left(25-16{i}^{2}\right)\right]$ $y=a\left[{x}^{2}-10x+\left(25-16\left(-1\right)\right)\right]$ $y=a\left({x}^{2}-10x+41\right)$ For simplicity, $leta=1$: $y={x}^{2}-10+41$