Question

Show that the roots of the equation ${x^2} - 2x + 3 = 0$ are imaginary.

Answer 1

Hint: The given equation is a quadratic equation so it will have two roots. The roots of this equation can be found out by using the formula method. The formula method states that, the roots of a quadratic equation of the form $a{x^2} + bx + c = 0$ is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where ${b^2} - 4ac = \Delta $

Complete step-by-step solution:
The given quadratic equation is ${x^2} - 2x + 3 = 0$.
By comparing it with the standard quadratic equation $a{x^2} + bx + c = 0$, we get $a = 1,b = - 2$ and $c = 3$.
Substituting these values in ${b^2} - 4ac = \Delta $,
$\Delta = {( - 2)^2} - 4(1)(3)$
Solving the square term and subtracting,
$\Delta = 4 - 12 = - 8$
$\therefore \Delta = - 8 < 0$
Since ${b^2} - 4ac < 0$, we can say that the roots of this equation are imaginary.
To find the roots we use the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and substitute all the values,
$x = \dfrac{{ - ( - 2) \pm \sqrt { - 8} }}{{2(1)}}$
Solving the brackets,
$x = \dfrac{{2 \pm \sqrt { - 8} }}{2}$
We can write $ - 8 = ( - 1)(8)$,
$x = \dfrac{{2 \pm \sqrt {( - 1)(8)} }}{2}$
We know the value of $\sqrt { - 1} = i$,
$\therefore x = \dfrac{{2 \pm i\sqrt 8 }}{2}$
Simplifying further,
$x = \dfrac{{2 \pm i\sqrt {2 \times 2 \times 2} }}{2}$
Taking a pair of $2$outside the root,
$x = \dfrac{{2 \pm 2i\sqrt 2 }}{2}$
Taking $2$ common from the numerator,
$x = \dfrac{{2(1 \pm i\sqrt 2 )}}{2}$
Cancelling $2$from numerator and denominator,
$x = 1 \pm i\sqrt 2 $
Therefore, we get the roots of ${x^2} - 2x + 3 = 0$ as $1 + i\sqrt 2 $ and $1 - i\sqrt 2 $ and both the roots are imaginary.
Hence, we found that the roots of ${x^2} - 2x + 3 = 0$ are imaginary.

Note: In the equation $a{x^2} + bx + c = 0$if ${b^2} - 4ac < 0$, then the roots of the quadratic equation are imaginary, if ${b^2} - 4ac > 0$, then the roots of the quadratic equation are real and unequal and if ${b^2} - 4ac = 0$, then the roots are real and equal of the quadratic equation.