Question

How do you find the complex roots of ${x^4} - 10{x^2} + 9 = 0$?

Answer 1

In this question we have to factorise the given polynomial using quadratic formula to find the complex roots, In the polynomial$a{x^2} + bx + c$, where "$a$", "$b$", and “$c$" are real numbers and the Quadratic Formula is formally stated as:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, and ${b^2} - 4ac$ is called the discriminant and if the discriminant is negative, i.e., if we have a negative under the radical then the roots of the quadratic equation will be complex conjugates.

Complete step by step solution:
Now the given quadratic equation is,
${x^4} - 10{x^2} + 9 = 0$,
As the equation has a degree 4 we will have four roots for the equation.
Imaginary or complex roots will occur when the value under the radical portion of the quadratic formula is negative. Notice that the value under the radical portion is represented by ${b^2} - 4ac$. So, if ${b^2} - 4ac$ is a negative value, the quadratic equation is going to have complex conjugate roots (containing "$i$ "s).
Now we will determine the discriminant i.e., ${b^2} - 4ac$,
Here $a = 1$,$b = - 10$,$c = 9$,
Now substituting the values in the discriminant we get,
$\Rightarrow {\left( { - 10} \right)^2} - 4\left( 1 \right)\left( 9 \right)$,
Now simplifying we get,
$\Rightarrow 100 - 36 = 64$,
So, ${b^2} - 4ac = 64 > 0$, so the given equation will have two real roots.
Now using the quadratic formula, here it is given by ${x^2} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
Here $a = 1$,$b = - 10$,$c = 9$,
Now substituting the values in the formula we get,
$\Rightarrow {x^2} = \dfrac{{ - \left( { - 10} \right) \pm \sqrt {{{\left( { - 10} \right)}^2} - 4\left( 1 \right)\left( 9 \right)} }}{{2\left( 1 \right)}}$,
Now simplifying we get,
$\Rightarrow {x^2} = \dfrac{{10 \pm \sqrt {100 - \left( {36} \right)} }}{2}$,
Now again simplifying we get,
$\Rightarrow {x^2} = \dfrac{{10 \pm \sqrt {64} }}{2}$,
Now taking the square root we get,
$\Rightarrow {x^2} = \dfrac{{10 \pm 8}}{2}$,
Now we get two values of $x$they are${x^2} = \dfrac{{10 + 8}}{2} = \dfrac{{18}}{2} = 9$and
,${x^2} = \dfrac{{10 - 8}}{2} = \dfrac{2}{2} = 1$
Now further simplifying taking out the square root we get,
$\Rightarrow x = \pm \sqrt 9$,and $x = \pm \sqrt 1$
Now simplifying we get, $x = \pm 3$,$x = \pm 1$.
So the zeros of the equation are $\pm 3$ and $\pm 1$, the equation will not have any complex roots.

Final Answer:
$\therefore$If we solve the given equation, i.e., ${x^4} - 10{x^2} + 9 = 0$, then four values of $x$ are $\pm 3$ and $\pm 1$, and the equation will not have any complex roots.

Note:
The discriminant, ${b^2} - 4ac$, offers valuable information about the "nature" of the roots of a quadratic equation where a, b and c are rational values. It quickly tells you if the equation has two real roots i.e., ${b^2} - 4ac > 0$, one real repeated root i.e., ${b^2} - 4ac = 0$ or two complex conjugate roots i.e., ${b^2} - 4ac < 0$. If you are trying to determine the "type" of roots of a quadratic equation we need not complete the entire quadratic formula. Simply look at the discriminant.