Question

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

Answer 1

`p=-4`
`q=13`

Explanation:

Suppose the roots of the general quadratic equation:

`ax^2+bx+c = 0`

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

`"sum of roots" \ \ \ \ \ \= alpha+beta = -b/a`
`"product of roots" = alpha beta \ \ \ \ = c /a`

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

`x^2+px+q = 0`

we know that:

`alpha+beta = -p/1 \ \ \ ` ; and `\ \ \ alpha beta = q/1`

And we can calculate:

`alpha + beta = 2+3i + 2-3i = 4 => p=-4`
`alpha beta = (2+3i)(2-3i) = 4+9 = 13 => q=13`

Answer 2

`p=-4, and, q=13.`

Explanation:

It is known that `2+3i` is a root of the Quadr. Eqn.`:x^2+px+q=0.`

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

`:. (2+3i)^2+p(2+3i)+q=0.`

`:. 4+12i+9i^2+2p+3ip+q=0.`

`:. 4+12i-9+2p+3ip+q=0.`

`:. (2p+q-5)+i(12+3p)=0.`

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

`2p+q=5......(1), and, 12+3p=0`

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

Enjoy Maths.!