Question

How do you solve \[{x^2} - 10x + 41 = 0\] ?

Answer 1

Hint: As the given equation is quadratic equation of the form \[a{x^2} + bx + c\] , in which x is an unknown term and a function is said to be differentiable if its derivative exists and hence to solve the given expression, we need to apply quadratic equation formula i.e., \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , in which we must substitute the values of a, b and c from the given equation.
Formula used:
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
In which, \[a,b,c \in R\] .

Complete step by step solution:
Let us write the given equation:
 \[{x^2} - 10x + 41 = 0\]
Since the equation cannot be easily factored, we must use the quadratic equation:
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
where a, b, c are from the standard form of a quadratic,
 \[f\left( x \right) = a{x^2} + bx + c\]
Hence, according to the given equation we have, \[a = 1\] , \[b = - 10\] , \[c = 41\] .
Now, substitute these values in the quadratic equation and simplify the terms:
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
 \[x = \dfrac{{ - \left( { - 10} \right) \pm \sqrt {{{\left( { - 10} \right)}^2} - 4\left( 1 \right)\left( {41} \right)} }}{{2\left( 1 \right)}}\]
We know that \[{\left( { - 10} \right)^2} = 100\] and hence evaluating the terms, we get:
 \[x = \dfrac{{10 \pm \sqrt {100 - 164} }}{2}\]
 \[x = \dfrac{{10 \pm \sqrt { - 64} }}{2}\]
Since we cannot square root a negative number, we use i to denote the imaginary unit, \[i = \sqrt { - 1} \] , and continue simplifying as:
 \[ \Rightarrow x = \dfrac{{10 \pm 8i}}{2}\]
As, there is a common denominator for the terms, hence we get:
 \[ \Rightarrow x = \dfrac{{10}}{2} \pm \dfrac{{8i}}{2}\]
 \[ \Rightarrow x = 5 \pm 4i\]
Therefore, our solutions are:
 \[x = 5 + 4i,5 - 4i\]
So, the correct answer is “ \[x = 5 + 4i,5 - 4i\] ”.

Note: Quadratic equations are the polynomial equations of degree two in one variable of type \[f\left( x \right) = a{x^2} + bx + c\] , hence we get the factors easily but here the given equation cannot be easily factored, so we need to apply the formula of Quadratic equation. And you must know the nature of roots i.e., from the formula we have \[D = {b^2} - 4ac\] , hence the nature of roots is:
If \[D > 0\] , roots are real and distinct (unequal), if \[D = 0\] roots are real and equal (coincident) and if \[D < 0\] , roots are imaginary and unequal.