# How do you solve the quadratic x^2-(4+i)x+9+7i=0 using any method?

Aug 15, 2016

$x = 1 + 3 i$ or $x = 3 - 2 i$

#### Explanation:

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The zeros of a quadratic in the form $a {x}^{2} + b x + c$ are given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In our example:

${x}^{2} - \left(4 + i\right) x + \left(9 + 7 i\right) = 0$

we have $a = 1$, $b = - \left(4 + i\right)$, $c = \left(9 + 7 i\right)$, so:

$x = \frac{\left(4 + i\right) \pm \sqrt{{\left(- \left(4 + i\right)\right)}^{2} - 4 \left(1\right) \left(9 + 7 i\right)}}{2 \cdot 1}$

$= \frac{\left(4 + i\right) \pm \sqrt{\left(16 + 8 i + {i}^{2}\right) - 36 - 28 i}}{2}$

$= \frac{\left(4 + i\right) \pm \sqrt{- 21 - 20 i}}{2}$

$= \frac{\left(4 + i\right) \pm i \sqrt{21 + 20 i}}{2}$

Sqaure root of a+bi

Note that I found in https://socratic.org/s/aw38evei

that the square roots of $a + b i$ are:

$\pm \left(\left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) + \left(\frac{b}{\left\mid b \right\mid} \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i\right)$

In particular, if $a + b i$ is in Q1, then its principal square root is

$\left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) + \left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i$

So we find:

$\sqrt{21 + 20 i}$

$= \left(\sqrt{\frac{\sqrt{{21}^{2} + {20}^{2}} + 21}{2}}\right) + \left(\sqrt{\frac{\sqrt{{21}^{2} + {20}^{2}} - 21}{2}}\right) i$

$= \left(\sqrt{\frac{\sqrt{441 + 400} + 21}{2}}\right) + \left(\sqrt{\frac{\sqrt{441 + 400} - 21}{2}}\right) i$

$= \left(\sqrt{\frac{\sqrt{841} + 21}{2}}\right) + \left(\sqrt{\frac{\sqrt{841} - 21}{2}}\right) i$

$= \left(\sqrt{\frac{29 + 21}{2}}\right) + \left(\sqrt{\frac{29 - 21}{2}}\right) i$

$= \left(\sqrt{\frac{50}{2}}\right) + \left(\sqrt{\frac{8}{2}}\right) i$

$= \left(\sqrt{25}\right) + \left(\sqrt{4}\right) i$

$= 5 + 2 i$

Conclusion

So:

$x = \frac{\left(4 + i\right) \pm i \sqrt{21 + 20 i}}{2}$

$= \frac{\left(4 + i\right) \pm i \left(5 + 2 i\right)}{2}$

$= \frac{\left(4 + i\right) \pm \left(- 2 + 5 i\right)}{2}$

$= \left\{\begin{matrix}\frac{2 + 6 i}{2} = 1 + 3 i \\ \frac{6 - 4 i}{2} = 3 - 2 i\end{matrix}\right.$

Aug 16, 2016

Factor $9 + 7 i$ then find a pair of factors summing to $4 + i$ to find zeros:

$1 + 3 i$ and $3 - 2 i$

#### Explanation:

${x}^{2} - \left(4 + i\right) x + \left(9 + 7 i\right) = 0$

Note that if this has roots ${x}_{1}$ and ${x}_{2}$ then:

${x}^{2} - \left(4 + i\right) x + \left(9 + 7 i\right)$

$= \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) = {x}^{2} - \left({x}_{1} + {x}_{2}\right) + {x}_{1} {x}_{2}$

So we want to find ${x}_{1}$, ${x}_{2}$ such that:

$\left\{\begin{matrix}{x}_{1} + {x}_{2} = 4 + i \\ {x}_{1} {x}_{2} = 9 + 7 i\end{matrix}\right.$

Gaussian integers

$4 + i$ and $9 + 7 i$ are both Gaussian integers, that is Complex numbers of the form $a + b i$ where $a$ and $b$ are integers.

What are the Gaussian integer factors of $9 + 7 i$?

$| | 9 + 7 i | | = \sqrt{{9}^{2} + {7}^{2}} = \sqrt{81 + 49} = \sqrt{130}$

Any Gaussian integer factors of $9 + 7 i$ must have norms with squares which are factors of $130$.

$130 = 2 \cdot 5 \cdot 13$

Up to a unit factor, the only Gaussian integer with norm $\sqrt{2}$ is $1 + i$.

$\frac{9 + 7 i}{1 + i} = \frac{\left(9 + 7 i\right) \left(1 - i\right)}{\left(1 + i\right) \left(1 - i\right)} = \frac{16 - 2 i}{2} = 8 - i$

Gaussian integers with norm $\sqrt{5}$ are $\pm 2 \pm i$ and $\pm 1 \pm 2 i$. Up to a unit factor, we only need to try $2 + i$ and $2 - i$:

$\frac{8 - i}{2 + i} = \frac{\left(8 - i\right) \left(2 - i\right)}{\left(2 + i\right) \left(2 - i\right)} = \frac{15 - 10 i}{5} = 3 - 2 i$

So:

$9 + 7 i = \left(1 + i\right) \left(2 + i\right) \left(3 - 2 i\right)$

Any unit ($\pm 1$ or $\pm i$) multiple of these factors or their products is also a factor of $9 + 7 i$.

Taking pairs of these factors we find:

$\left(1 + i\right) \left(2 + i\right) = 1 + 3 i \text{ }$ with co-factor $\left(3 - 2 i\right)$

$\left(2 + i\right) \left(3 - 2 i\right) = 8 - i \text{ }$ with co-factor $\left(1 + i\right)$

$\left(1 + i\right) \left(3 - 2 i\right) = 6 - 5 i \text{ }$ with co-factor $\left(2 + i\right)$

The first of these sums to $4 + i$ as required:

$\left(1 + 3 i\right) + \left(3 - 2 i\right) = 4 + i$

So our zeros are $1 + 3 i$ and $3 - 2 i$