Question

If one root of the equation $a{x}^2-6x+c+9=0$ $\left(a,c\in R,a\ne 0\right)$ is $3-5i$ then $a=\_\_\_c$

• A. $1,25$
• B. $-1,25$
• C. $1,-25$
• D. $-1,-25$

Answer 1

The correct option is A

$1,25$

Explanation for the correct option

Step 1: Information required for the solution

We know that the quadratic equation of the forms $x^2+ax+b=0$ has its sum of the roots equal to the coefficient of the $x$ that is $a$ and product of the roots is equal to the constant $b$.

To convert the given equation in the form $x^2+ax+b=0$ we need to divide the equation $a{x}^2-6x+c+9=0$ by $a$.

The equation will become $x^2-\frac{6}{a}+\frac{c+9}{a}=0$.

Let the roots of the equation be $\alpha \text{and}\beta$ then the two equations will be,

$\alpha +\beta =\frac{6}{a}\dots \left(1\right)$

$\alpha \beta =\frac{c+9}{a}\dots \left(2\right)$

Step 2: Calculation of the value

Since one root of the equation is $3-5i$ let this be equal to $\alpha$. The other root will be $3+5i$ which is equal to $\beta$.

$\begin{array}{rcl}\therefore \alpha +\beta & =& 3+5i+3-5i\\ & =& 6\end{array}$

From equation $\left(1\right)$,

$\begin{array}{rcl}\therefore \frac{6}{a}& =& 6\\ & \Rightarrow & a=1\end{array}$

Now, the product of the roots will be

$\begin{array}{rcl}\therefore \alpha \beta & =& \left(3-5i\right)\left(3+5i\right)\\ & =& 9-\left(25\right)\left(-1\right)\\ & =& 34\end{array}$

From equation $\left(2\right)$,

$\begin{array}{rcl}\therefore \frac{c+9}{a}& =& 34\\ & \Rightarrow & c+9=34\\ & \Rightarrow & c=25\end{array}$

Therefore, the value of $a=1$ and $c=25$.

Hence, the correct option is (A).