One root of $x^2 + px + q = 0$, for $p, q$ real, is $x = 2 + 3i$. How do you find $p$ and $q$?
What I have tried so far:
$(2+3i)(2-3i)$
$=4-6i+6i-9i^2$
$=4+9 = 13$
Therefore $q$ is $13$
But I don't know how to find the value of $p$
Any help would be appreciated!
On
$p$ and $q$ are not unique as the same polynomial multiplied by any nonzero constant has the same roots. Other than that Siong Thye Goh showed how it's done.
On
The roots must occur in conjugate pairs (why?), so the polynomial must actually be $$(x-2-3i)(x-2+3i)$$ $$=x^2 -4x +13$$
because $(x-z)(x-\bar{z}) = x^2 -2 \operatorname{Re} z +|z|^2$.
On
This argument starts from less information. Suppose we didn't know that the "other root" of the quadratic equation (call it $ \ r \ $ ) is the complex conjugate of $ \ 2 + 3i \ \ , $ but we wish the coefficients $ \ p \ $ and $ \ q \ $ to be real numbers. The polynomial would factor as $ \ ( x - 2 - 3i ) · ( x - r ) \ = \ 0 \ \ , $ giving us $$ x^2 \ + \ px \ + \ q \ \ = \ \ x^2 \ - \ (2 + 3i + r)·x \ + \ (2 + 3i)·r \ \ = \ \ 0 \ \ . $$ We see from this that $ \ r \ $ cannot simply be a real number because both $ \ p \ = \ -(2 + 3i + r) \ $ and $ \ q \ = \ (2 + 3i)·r \ $ would be left with imaginary parts.
We can make $ \ p \ $ a real number by choosing $ \ r \ = \ a - 3i \ \ , $ so that $ \ p \ $ becomes equal to $ \ -(2 + a) \ \ . $ As for the other coefficient, we find $ \ q \ = \ (2 + 3i)·(a - 3i) \ = \ (2a + 9) \ + \ (3a - 6)·i \ \ . $ We can make this a real number by setting $ \ 3a - 6 \ = \ 0 \ \Rightarrow \ a \ = \ 2 \ \ . $ From this, we have $$ r \ = \ 2 - 3i \ \ , \ \ p \ = \ -(2 + 2) \ = \ -4 \ \ , \ \ q \ = \ (2·2 + 9) \ + \ 0·i \ = \ 13 \ \ , $$ and the quadratic equation, $ \ x^2 - 4x + 13 \ = \ 0 \ \ . $
Note that if $\alpha$ and $\beta$ are the roots, then we have
$(x-\alpha)(x-\beta)=0$
$$x^2-(\alpha+\beta) x+\alpha \beta = 0$$
Since your equation is monic,
$$p=-(2+3i+2-3i)=-4$$