Question

If one of the roots of the equation $2{x^2} - 6x + k = 0$ is $(\alpha + 5i)/2$ , then the values of $\alpha $ and $k$ are
A.$\alpha = 3,k = 8$
B.$\alpha = \dfrac{3}{2},k = 17$
C.$\alpha = 2,k = 16$
D.$\alpha = 3,k = 17$

Answer 1

Hint: Here we will standard quadratic equation and the sum of the roots and product of the roots formula. Follow step by step approach first sum of the roots, product of roots and then place the values in the given conditions and simplify accordingly.

Complete step-by-step answer:
Let us consider the general form of the quadratic equation.
$a{x^2} + bx + c = 0$
Compare the standard equation with the given equation – $2{x^2} - 6x + k = 0$ .....(A)
$
   \Rightarrow a = 2 \\
   \Rightarrow b = - 6 \\
   \Rightarrow c = k \\
 $
Given that one of the root is $ = \dfrac{{\alpha + 5i}}{2}$
Since the conjugate roots are always in pairs.
We get other root $ = \dfrac{{\alpha - 5i}}{2}$
Now, sum of the root is $ = \dfrac{{ - b}}{a}$
Place values by using equation (A)
Sum of the root is $ = \dfrac{{ - ( - 6)}}{2}$
Product of minus into minus gives plus and simplify
Sum of the root is $ = \dfrac{6}{2} = 3$
Now, place the values of roots in the above equation –
$\dfrac{{\alpha + 5i}}{2} + \dfrac{{\alpha - 5i}}{2} = 3$
Simplify the above equation –
$\dfrac{{\alpha + 5i + \alpha - 5i}}{2} = 3$
Like terms with the opposite sign cancel each other and same signs add.
\[
  \dfrac{{\alpha + \alpha }}{2} = 3 \\
   \Rightarrow \dfrac{{2\alpha }}{2} = 3 \\
 \]
Same terms from the numerator and the denominator cancel each other.
$ \Rightarrow \alpha = 2\,{\text{ }}....{\text{ (B)}}$
Now, We know that product of roots is $\dfrac{c}{a} = \dfrac{k}{2}$
Place values of the given roots –
 $\left( {\dfrac{{\alpha + 5i}}{2}} \right)\left( {\dfrac{{\alpha - 5i}}{2}} \right) = \dfrac{k}{2}$
Simplify the above equation- by using the difference of two squares. $(a + b)(a - b) = {a^2} - {b^2}$
$\left( {\dfrac{{{\alpha ^2} - {{(5i)}^2}}}{4}} \right) = \dfrac{k}{2}$
Place the value of $\alpha $.
$ \Rightarrow \left( {\dfrac{{{3^2} - (25{i^2})}}{4}} \right) = \dfrac{k}{2}$
Also, ${i^2} = ( - 1)$
$ \Rightarrow \left( {\dfrac{{9 - (25( - 1))}}{4}} \right) = \dfrac{k}{2}$
Product of minus term and minus terms gives positive terms.
$ \Rightarrow \left( {\dfrac{{9 + (25)}}{4}} \right) = \dfrac{k}{2}$
Simplify the above equation-
$ \Rightarrow \left( {\dfrac{{34}}{4}} \right) = \dfrac{k}{2}$
Do-cross multiplication and simplify –
$ \Rightarrow \left( {\dfrac{{34}}{4}} \right) \times 2 = k$
Take common factors from the numerator and the denominator and remove them.
$
   \Rightarrow 17 = k \\
   \Rightarrow k = 17 \\
 $
Hence, from the given multiple choices – the option D is the correct answer.

Note: A quadratic equation is an equation of second degree, it means at least one of the terms is squared. Standard equation is $a{x^2} + bx + c = 0$ where a,b,and c are constant and “a” can never be zero and “x” is unknown. Always remember the standard equation and formula for the sum and product of the roots properly. Follow the given data and conditions carefully and simplify.