If \[2 + 3i\] is one of the root of the equation \[2{x^3} - 9{x^2} + kx - 13 = 0,k \in \Re \], then the real root of this equation:
A.Exist and equal to \[\dfrac{1}{2}\]
B.Does not exist
C.Exist and equal to \[1\]
D.Exist and equal to \[ - \dfrac{1}{2}\]
Answer
663.3k+ views
Hint- If \[a + ib\] is an imaginary root of an equation\[p{x^3} + q{x^2} + rx + s = 0\], then \[a - ib\] will be another imaginary roots of this equation.
Again, let us consider, \[\alpha ,\beta ,\gamma \] to be three roots of the above equation. Then, by the relation between roots and coefficients we get,
\[
\alpha + \beta + \gamma = \dfrac{{ - q}}{p} \\
\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{r}{p} \\
\alpha \beta \gamma = \dfrac{{ - s}}{p} \\
\]
Complete step by step answer:
Since, the highest power of the given equation is three and one root are imaginary then, it has another imaginary root and a real root.
According to the problem, \[2 + 3i\] is one of the roots of the equation \[2{x^3} - 9{x^2} + kx - 13 = 0,k \in \Re \], then \[2 - 3i\] is another roots of the given equation.
Let us consider, \[\alpha ,\beta ,\gamma \] be three roots of the above equation, where,
\[\alpha = 2 + 3i,\beta = 2 - 3i\] and \[\gamma \] be the real root.
Form the relation between the roots and coefficients we get,
\[\alpha + \beta + \gamma = \dfrac{9}{2}\]
Substitute the value of \[\alpha = 2 + 3i,\beta = 2 - 3i\] in the above equation we get,
\[2 + 3i + 2 - 3i + \gamma = \dfrac{9}{2}\]
Now let us solve the equation to find \[\gamma \] , we get,
\[4 + \gamma = \dfrac{9}{2}\]
Now let us again solve again we get,
\[\gamma = \dfrac{9}{2} - 4 = \dfrac{1}{2}\]
Hence, We have found that the real root exists.
Therefore, The real root exists and is equal to \[\dfrac{1}{2}\].
The correct option is (A) Exist and equal to \[\dfrac{1}{2}\].
Note: For a cubic equation, it has either three real roots or two imaginary and one real root. If \[a + ib\] is an imaginary root of the cubic equation, then its reciprocal \[a - ib\] will be another imaginary root of the same equation.
Again, let us consider, \[\alpha ,\beta ,\gamma \] to be three roots of the above equation. Then, by the relation between roots and coefficients we get,
\[
\alpha + \beta + \gamma = \dfrac{{ - q}}{p} \\
\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{r}{p} \\
\alpha \beta \gamma = \dfrac{{ - s}}{p} \\
\]
Complete step by step answer:
Since, the highest power of the given equation is three and one root are imaginary then, it has another imaginary root and a real root.
According to the problem, \[2 + 3i\] is one of the roots of the equation \[2{x^3} - 9{x^2} + kx - 13 = 0,k \in \Re \], then \[2 - 3i\] is another roots of the given equation.
Let us consider, \[\alpha ,\beta ,\gamma \] be three roots of the above equation, where,
\[\alpha = 2 + 3i,\beta = 2 - 3i\] and \[\gamma \] be the real root.
Form the relation between the roots and coefficients we get,
\[\alpha + \beta + \gamma = \dfrac{9}{2}\]
Substitute the value of \[\alpha = 2 + 3i,\beta = 2 - 3i\] in the above equation we get,
\[2 + 3i + 2 - 3i + \gamma = \dfrac{9}{2}\]
Now let us solve the equation to find \[\gamma \] , we get,
\[4 + \gamma = \dfrac{9}{2}\]
Now let us again solve again we get,
\[\gamma = \dfrac{9}{2} - 4 = \dfrac{1}{2}\]
Hence, We have found that the real root exists.
Therefore, The real root exists and is equal to \[\dfrac{1}{2}\].
The correct option is (A) Exist and equal to \[\dfrac{1}{2}\].
Note: For a cubic equation, it has either three real roots or two imaginary and one real root. If \[a + ib\] is an imaginary root of the cubic equation, then its reciprocal \[a - ib\] will be another imaginary root of the same equation.
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