Question

How do you solve ${x^2} - 2x + 10 = 0?$

Answer 1

Hint: This problem deals with solving a quadratic equation. To solve any quadratic equation, first put all the terms on one side of the equal sign, leaving zero on the other side. Then factorize the expression. Now set each factor equal to zero. Solve each of these equations. Check by inserting your answer in the original equation. This is one way of solving the quadratic equation, another is by using the formula.
Given a quadratic equation: $a{x^2} + bx + c = 0$, then the roots of the equation is given by:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
Given an equation which is an expression in quadratic.
That is a quadratic equation is given, which is ${x^2} - 2x + 10 = 0$.
To solve a quadratic expression is the same as finding the roots of the quadratic equation.
Consider the given quadratic equation, as given below:
$ \Rightarrow {x^2} - 2x + 10 = 0$
The roots of the above quadratic equation are obtained by applying the formula of roots of the quadratic equation, as given below:
$ \Rightarrow x = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( {10} \right)} }}{{2\left( 1 \right)}}$
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 - 40} }}{2}$
Simplifying the above expression inside the under root, gives the following expression:
$ \Rightarrow x = \dfrac{{2 \pm \sqrt { - 36} }}{2}$
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {36( - 1)} }}{2} = \dfrac{{2 \pm \sqrt { - 1} \left( {\sqrt {36} } \right)}}{2}\]
We know that $\sqrt { - 1} = i$, substitute this in the above expression, as given below:
\[ \Rightarrow x = \dfrac{{2 \pm i\sqrt {36} }}{2}\]
We know that the square root of 36 is 6, which is $\sqrt {36} = 6$, substituting this in the equation above,
\[ \Rightarrow x = \dfrac{{2 \pm i6}}{2}\]
\[ \Rightarrow x = 1 \pm 3i\]
$\therefore x = 1 + 3i;$ and $x = 1 - 3i$
Hence the roots of the quadratic equation are complex and are given by:
$ \Rightarrow x = 1 \pm 3i$

The solutions of the quadratic equation ${x^2} - 2x + 10 = 0$ are $1 \pm 3i$.

Note: Note that in any quadratic equation $a{x^2} + bx + c = 0$, here in this ${b^2} - 4ac$ is called as the determinant. Generally there are 3 general cases of the discriminant.
If ${b^2} - 4ac > 0$, then the quadratic equation has real and distinct roots.
If ${b^2} - 4ac = 0$, then the quadratic equation has real and equal roots.
If ${b^2} - 4ac < 0$, then the quadratic equation has no real roots but has complex conjugate roots.