Question

The quadratic equation of real coefficients, whose one root is $7+5i$, will be
A. $x^{2}+14x+74=0$
B. $x^{2}-14x-74=0$
C. $x^{2}+14x-74=0$
D. $x^{2}-14x+74=0$

Answer 1

Hint:The quadratic equation has real coefficients; it will be greater than 0 whose one of the roots is already given. We need to assume the other root is a conjugate of the first one. Then, we need to find the sum of roots as well as the product of the roots. Substituting their value in the formula used we will get the quadratic equation

Formula Used:
$x^{2}-\left ( \alpha +\beta \right )x+\alpha \beta= 0 $
$i=\sqrt{-1}$

Complete step by step Solution:
Since the coefficients of a quadratic equation are real, let us assume the other root must be the conjugate of the first root.
Let $\alpha $ and $\beta $ be the two roots of the required quadratic equation.
Thus,
$\alpha =7+5i$
$\beta =7+5i$
Therefore,
$\alpha +\beta =7+5i+7-5i$
$\alpha +\beta =14$
Similarly,
$\alpha \beta =\left ( 7+5i \right )\left ( 7-5i \right )$
On further solving, we get
$\alpha \beta =\left ( 7 \right )^{2}-\left ( 5i \right )^{2}=49-25\left ( i^{2} \right )$
Now, substitute the value of i

$\alpha \beta $=49+25=74
Now, the required quadratic equation can be found by using the below-given formula
$x^{2}-\left ( \alpha +\beta \right )x+\alpha \beta $
Now, substituting the values of the sum of roots and the product of the roots in the above equation, we get
$x^{2}-14x+74=0$
Hence, the quadratic equation whose roots are $\alpha $ and $\beta $ is
$x^{2}-14x+74=0$

Hence, the correct option is (B).


Note: Students need to ensure the bother root is the conjugate value of the first root which is the key to finding the required quadratic equation. In mathematics, a pair of binomials with identical phrases that part opposite arithmetic operators amid these similar terms are referred to as conjugates. In mathematics, a couple of binomials with exact phrases that part opposite arithmetic operators amid these similar terms are referred to as conjugates.