Question

How do you solve ${x^2} - 2x + 9 = 0$

Answer 1

Hint: This problem comes under solving quadratic equations in algebra. The general form of the quadratic equation is $a{x^2} + bx + c = 0$. There are methods to solve quadratic equations; they are factorisation method, formula method and complete square root method. Here we use formula methods because the values are a little complicated. If the values are small we use factorisation methods to solve. Here there will be a formula on comparing the general form of quadratic equation with equation and substitute the values in the formula we will get required roots value.

Formula used: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
Let us consider the quadratic equation ${x^2} - 2x + 9 = 0$
Now on comparing coefficient of ${x^2}$, $x$ and constant term in general form of quadratic equation $a{x^2} + bx + c = 0$ with ${x^2} - 2x + 9 = 0$ , we get
$a = 1,b = - 2,c = 9$
Now substitute the values of a, b, c in the formula mentioned in formula used, we get
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(1)(9)} }}{{2(1)}}$
On simplification we get
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 - 36} }}{2}$
Let us subtract the term and we get
$ \Rightarrow x = \dfrac{{2 \pm \sqrt { - 32} }}{2}$
So we get,
$ \Rightarrow \Delta = {b^2} - 4ac$
$\therefore \Delta < 0$ the roots are imaginary.
Since the roots are imaginary we use complex number $\sqrt { - 1} = i$
Now simplifying the roots to complex number, we get
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {32} \sqrt { - 1} }}{2}$
On rewriting we get
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {32} i}}{2}$
On multiply we get
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 \times 8} i}}{2}$
On splitting the term and we get
$ \Rightarrow x = \dfrac{{2 \pm 2\sqrt 8 i}}{2}$
Taking 2 as common and we get
$ \Rightarrow x = \dfrac{{2(1 \pm \sqrt 8 i)}}{2}$
By cancelling, we get
$ \Rightarrow x = 1 \pm \sqrt 8 i$

Therefore the roots are $x = 1 + \sqrt 8 i,1 - \sqrt 8 i$

Note: Here we find the roots of complex numbers since the roots are imaginary. We can determine the nature of roots by delta. \[\Delta = {b^2} - 4ac\],
If $\Delta = 0,$ the roots are real and equal
If $\Delta > 0,$ the roots are real and unequal
$\Delta < 0,$ The roots are imaginary. By this we can find the nature of roots. The complex roots of the quadratic equation are represented as \[a + ib\] and \[a - ib\]. Thus we find the solution of this model.