We will discuss about the complex roots of a quadratic equation.
In a quadratic equation with real coefficients has a complex root α + iβ then it has also the conjugate complex root α  iβ.
Proof:
To prove the above theorem let us consider the quadratic equation of the general form:
ax\(^{2}\) + bx + c = 0 where, the coefficients a, b and c are real.
Let α + iβ (α, β are real and i = √1) be a complex root of equation ax\(^{2}\) + bx + c = 0. Then the equation ax\(^{2}\) + bx + c = 0 must be satisfied by x = α + iβ.
Therefore,
a(α + iβ)\(^{2}\) + b(α + iβ) + c = 0
or, a(α\(^{2}\)  β\(^{2}\) + i ∙ 2 αβ) + bα + ibβ + c = 0, (Since, i\(^{2}\) = 1)
or, aα\(^{2}\)  aβ\(^{2}\) + 2iaαβ + bα + ibβ + c = 0,
or, aα\(^{2}\)  aβ\(^{2}\) + bα + c + i(2aαβ + bβ) = 0,
Therefore,
aα\(^{2}\)  aβ\(^{2}\) + bα + c = 0 and 2aαβ + bβ = 0
Since, p + iq = 0 (p, q are real and i = √1) implies p = 0 and q = 0]
Now substitute x by α  iβ in ax\(^{2}\) + bx + c we get,
a(α  iβ)\(^{2}\) + b(α  iβ) + c
= a(α\(^{2}\)  β\(^{2}\)  i ∙ 2 αβ) + bα  ibβ + c, (Since, i\(^{2}\) = 1)
= aα\(^{2}\)  aβ\(^{2}\)  2iaαβ + bα  ibβ + c,
= aα\(^{2}\)  aβ\(^{2}\) + bα + c  i(2aαβ + bβ)
= 0  i ∙ 0 [Since, aα\(^{2}\)  aβ\(^{2}\) + bα + c = 0 and 2aαβ + bβ = 0]
= 0
Now we clearly see that the equation ax\(^{2}\) + bx + c = 0 is satisfied by x = (α  iβ) when (α + iβ) is a root of the equation. Therefore, (α  iβ) is the other complex root of the equation ax\(^{2}\) + bx + c = 0.
Similarly, if (α  iβ) is a complex root of equation ax\(^{2}\) + bx + c = 0 then we can easily proved that its other complex root is (α + iβ).
Thus, (α + iβ) and (α  iβ) are conjugate complex roots. Therefore, in a quadratic equation complex or imaginary roots occur in conjugate pairs.
Solved example to find the imaginary roots occur in conjugate pairs of a quadratic equation:
Find the quadratic equation with real coefficients which has 3  2i as a root (i = √1).
Solution:
According to the problem, coefficients of the required quadratic equation are real and its one root is 3  2i. Hence, the other root of the required equation is 3  2i (Since, the complex roots always occur in pairs, so other root is 3 + 2i.
Now, the sum of the roots of the required equation = 3  2i + 3 + 2i = 6
And, product of the roots = (3 + 2i)(3  2i) = 3\(^{2}\)  (2i)\(^{2}\) = 9  4i\(^{2}\) = 9 4(1) = 9 + 4 = 13
Hence, the equation is
x\(^{2}\)  (Sum of the roots)x + product of the roots = 0
i.e., x\(^{2}\)  6x + 13 = 0
Therefore, the required equation is x\(^{2}\)  6x + 13 = 0.
`11 and 12 Grade Math
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