
If one of the roots of the equation \[{x^2} + \left( {1 - 3i} \right)x - 2\left( {1 + i} \right) = 0\] is \[ - 1 + i\] then the other root is
A) \[ - 1 - i\]
B) \[\dfrac{{ - 1 - i}}{2}\]
C) i
D) 2i
Answer
519.9k+ views
Hint: Given equation is definitely of quadratic form but along with that it has a touch of complex numbers that are the roots of the equation are complex numbers. In the case above one root is given but another is to be found. We will use the relation between the roots to find the other root.
Complete step by step answer:
If \[a{x^2} + bx + c = 0\] is the general quadratic equation and its roots are \[\alpha \& \beta \] then the relation of the roots is given as,
\[\alpha + \beta = \dfrac{{ - b}}{a}\] and \[\alpha .\beta = \dfrac{c}{a}\] where a, b are the coefficients and c is the constant term.
Now we will use the first relationship here.
\[\alpha + \beta = \dfrac{{ - b}}{a}\]
Putting the values from the equation given \[{x^2} + \left( {1 - 3i} \right)x - 2\left( {1 + i} \right) = 0\] on comparing with \[a{x^2} + bx + c = 0\]. Then we have
\[\alpha + \left( { - 1 + i} \right) = \dfrac{{ - \left( {1 - 3i} \right)}}{1}\]
We will separate the unknown and the constants,
\[\alpha = - 1 + 3i - i + 1\]
On solving we get,
\[\alpha = 2i\]
Therefore, the other root of the given Quadratic equation is $2i$. Thus, option (D) is the correct.
Note:
Note that, we can solve this by using quadratic formula also but that becomes too long and time consuming. This answer is very feasible to find with the help of the relation between the roots.
Also note that in the relation \[\alpha + \beta = \dfrac{{ - b}}{a}\] the minus sign is very important because that can change the answer also. We can sue the relation that has product of the roots if we want.
Complete step by step answer:
If \[a{x^2} + bx + c = 0\] is the general quadratic equation and its roots are \[\alpha \& \beta \] then the relation of the roots is given as,
\[\alpha + \beta = \dfrac{{ - b}}{a}\] and \[\alpha .\beta = \dfrac{c}{a}\] where a, b are the coefficients and c is the constant term.
Now we will use the first relationship here.
\[\alpha + \beta = \dfrac{{ - b}}{a}\]
Putting the values from the equation given \[{x^2} + \left( {1 - 3i} \right)x - 2\left( {1 + i} \right) = 0\] on comparing with \[a{x^2} + bx + c = 0\]. Then we have
\[\alpha + \left( { - 1 + i} \right) = \dfrac{{ - \left( {1 - 3i} \right)}}{1}\]
We will separate the unknown and the constants,
\[\alpha = - 1 + 3i - i + 1\]
On solving we get,
\[\alpha = 2i\]
Therefore, the other root of the given Quadratic equation is $2i$. Thus, option (D) is the correct.
Note:
Note that, we can solve this by using quadratic formula also but that becomes too long and time consuming. This answer is very feasible to find with the help of the relation between the roots.
Also note that in the relation \[\alpha + \beta = \dfrac{{ - b}}{a}\] the minus sign is very important because that can change the answer also. We can sue the relation that has product of the roots if we want.
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