# The quadratic equation  x^2+px+q = 0 has a complex root 2+3i. Find p and q?

Apr 1, 2017

$p = - 4$
$q = 13$

#### Explanation:

Suppose the roots of the general quadratic equation:

$a {x}^{2} + b x + c = 0$

are $\alpha$ and $\beta$ , then using the root properties we have:

$\text{sum of roots} \setminus \setminus \setminus \setminus \setminus \setminus = \alpha + \beta = - \frac{b}{a}$
$\text{product of roots} = \alpha \beta \setminus \setminus \setminus \setminus = \frac{c}{a}$

Complex roots always appear in conjugate pairs, so if one root of the given quadratic is $\alpha = 2 + 3 i$ then the other root is $\beta = 2 - 3 i$

${x}^{2} + p x + q = 0$

we know that:

$\alpha + \beta = - \frac{p}{1} \setminus \setminus \setminus$ ; and $\setminus \setminus \setminus \alpha \beta = \frac{q}{1}$

And we can calculate:

$\alpha + \beta = 2 + 3 i + 2 - 3 i = 4 \implies p = - 4$
$\alpha \beta = \left(2 + 3 i\right) \left(2 - 3 i\right) = 4 + 9 = 13 \implies q = 13$

Apr 1, 2017

$p = - 4 , \mathmr{and} , q = 13.$

#### Explanation:

It is known that $2 + 3 i$ is a root of the Quadr. Eqn.$: {x}^{2} + p x + q = 0.$

Therefore, $x = 2 + 3 i$ must satisfy the eqn.

$\therefore {\left(2 + 3 i\right)}^{2} + p \left(2 + 3 i\right) + q = 0.$

$\therefore 4 + 12 i + 9 {i}^{2} + 2 p + 3 i p + q = 0.$

$\therefore 4 + 12 i - 9 + 2 p + 3 i p + q = 0.$

$\therefore \left(2 p + q - 5\right) + i \left(12 + 3 p\right) = 0.$

Comparing the Real and Imaginary Parts of both sides, we get,

$2 p + q = 5. \ldots . . \left(1\right) , \mathmr{and} , 12 + 3 p = 0$

rArr p=-4, &, q=5-2p=5-2(-4)=13, as Respected Steve Sir, has alreay derived.

Enjoy Maths.!