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If \[\alpha ,\beta \in R\] such that \[1 - 2i\] (here \[{i^2} = - 1\])is a root of \[{z^2} + \alpha z + \beta = 0\], the what is the value of \[\alpha - \beta \]?
A.7
B.-3
C.3
D.-7

Answer
VerifiedVerified
162k+ views
Hint: In solving the above question, first we will us the fact that the complex roots always come in pairs, which means actually that the non-real complex roots or imaginary roots occur as conjugate pairs, and the coefficients of the polynomial equation should be real.
For example, if the one complex root of the equation \[a{x^2} + bx + c = 0\] is \[p + iq\], then the other root will always be \[p - iq\], and then by simplifying by using the sum of roots and product of the roots facts we will get the desired result.
Formula used :
We will use the fact that for the equation \[a{x^2} + bx + c = 0\],
Sum of roots \[ = \dfrac{{ - b}}{a}\] and
Product of the roots \[ = \dfrac{c}{a}\].
Complete step by step solution:
Given \[\alpha ,\beta \in R\] such that \[1 - 2i\]is a root of \[{z^2} + \alpha z + \beta = 0\],
Now here one complex root is \[1 - 2i\], then by using the complex root facts, the other root will be \[1 + 2i\],
So, the roots of the equation given i.e.,\[{z^2} + \alpha z + \beta = 0\] are \[1 - 2i\] and \[1 + 2i\],
Now using fact of sum of the roots and product of the roots, we will get,
For the equation \[a{x^2} + bx + c = 0\], sum of roots \[ = \dfrac{{ - b}}{a}\] and Product of the roots \[ = \dfrac{c}{a}\].
By comparing the above equation with the given equation i.e., \[{z^2} + \alpha z + \beta = 0\],
\[a = 1,b = \alpha ,c = \beta \],
Sum of the roots\[ = \dfrac{{ - \alpha }}{1}\]\[ = - \alpha \],
And product of roots \[ = \dfrac{\beta }{1}\]\[ = \beta \],
Now here we know that roots of the given equation are \[1 - 2i\] and \[1 + 2i\],
Now we will equate the like terms we will get,
\[ \Rightarrow - \alpha = 1 - 2i + 1 + 2i\]
Now by simplifying we will get,
\[ \Rightarrow - \alpha = 2\]
Now we will further simplify we will get,
\[ \Rightarrow \alpha = - 2\]
Now we will equate the other term,
\[ \Rightarrow \beta = \left( {1 - 2i} \right)\left( {1 + 2i} \right)\]
Now we will multiply the terms, we will get,
\[ \Rightarrow \beta = 1 + 2i - 2i - 4{i^2}\]
Now will simplify, we will get,
\[ \Rightarrow \beta = 1 - 4\left( { - 1} \right)\]( \[{i^2} = - 1\])
Now we will further simplify, we will get,
\[ \Rightarrow \beta = 1 + 4\],
Now we will add the terms , we will get,
\[ \Rightarrow \beta = 5\]
Now we will substitute the values of \[\alpha ,\beta \] in the expression \[\alpha - \beta \], we will get,
\[ \Rightarrow \alpha - \beta = - 2 - 5\]
Now finally we will get,
\[ \Rightarrow \alpha - \beta = - 7\]
The correct option is D
Note: Another method to solve the question is,
As the root of the given equation \[{z^2} + \alpha z + \beta = 0\]is\[1 - 2i\], substitute the root in the equation we will get,
\[ \Rightarrow {\left( {1 - 2i} \right)^2} + \alpha \left( {1 - 2i} \right) + \beta = 0\]
\[ \Rightarrow 1 + 4{i^2} - 2i + \alpha - 2i\alpha + \beta = 0\]
\[ \Rightarrow \left( {\alpha + \beta - 3} \right) - i\left( {4 + 2\alpha } \right) = 0\]
\[ \Rightarrow \left( {\alpha + \beta - 3} \right) - i\left( {4 + 2\alpha } \right) = 0 + i0\]
Now we will equate each term to 0,
\[ \Rightarrow \alpha + \beta - 3 = 0\] and \[4 + 2\alpha = 0\]
Now solving the second equation, we will get,
\[ \Rightarrow 2\alpha = - 4\]
\[ \Rightarrow \alpha = - \dfrac{4}{2}\]
\[ \Rightarrow \alpha = - 2\]
Now substituting the value of \[\alpha \]in the first equation, we will get,
\[ \Rightarrow - 2 + \beta - 3 = 0\]
\[ \Rightarrow \beta = 2 + 3\]
\[ \Rightarrow \beta = 5\]
Now we will substitute the values of \[\alpha ,\beta \] in the expression \[\alpha - \beta \], we will get,
\[ \Rightarrow \alpha - \beta = - 2 - 5\]
Now finally we will get,
\[ \Rightarrow \alpha - \beta = - 7\]