Question

Solve the equation: $- {x^2} + x - 2 = 0$

Answer 1

Here we are asked to solve the given equation which is a quadratic equation. A quadratic equation is an equation that has only one unknown variable with the highest degree two. The number of the roots of an equation depends on its degree, that is if the degree of the equation is two then it will have two roots. Here we are also given a quadratic equation so we will get two roots for this equation. The roots of the quadratic equation can be found by using the formula given in the following section.

Formula: If we have an quadratic equation in the form $a{x^2} + bx + c = 0$ then its roots can be found by using the formula: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
It is given that $- {x^2} + x - 2 = 0$ we aim to solve this equation, that is we have to find its roots.
We know that the number of roots of an equation is equal to its degree. Here the degree of the given equation is two thus this equation will have two roots.
The roots of a quadratic equation can be found by using the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a$ - coefficient of the term ${x^2}$, $b$ - coefficient of the term $x$, and $c$ - constant term.
First, let us collect the required terms for the formula from the given quadratic equation to solve it.
From the given equation $- {x^2} + x - 2 = 0$, we have $a = - 1$, $b = 1$, and $c = - 2$.
Now let us substitute the above values in the quadratic roots formula.
$x = \dfrac{{ - \left( 1 \right) \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( { - 1} \right)\left( { - 2} \right)} }}{{2\left( { - 1} \right)}}$
On simplifying this we get
$x = \dfrac{{ - 1 \pm \sqrt {1 - 8} }}{{ - 2}}$
On further simplification we get
$x = \dfrac{{ - 1 \pm \sqrt { - 7} }}{{ - 2}}$
$= \dfrac{{ - 1 \pm i\sqrt 7 }}{{ - 2}}$
$\Rightarrow x = \dfrac{{ - 1 + i\sqrt 7 }}{{ - 2}}$ and $\Rightarrow x = \dfrac{{ - 1 - i\sqrt 7 }}{{ - 2}}$
On solving the above, we get
$\Rightarrow x = \dfrac{{1 - i\sqrt 7 }}{2}$ and $\Rightarrow x = \dfrac{{1 + i\sqrt 7 }}{2}$
Thus, we got the roots of the given quadratic equation that is $x = \dfrac{{1 - i\sqrt 7 }}{2}$ and $x = \dfrac{{1 + i\sqrt 7 }}{2}$

Note: In the above problem, we got complex roots that is $x = \dfrac{{1 - i\sqrt 7 }}{2}$ and $x = \dfrac{{1 + i\sqrt 7 }}{2}$ , this is because in the calculation of finding the roots, we got an negative number inside the square root, the square root of negative will give the imaginary number $i$ to be more clear, $\sqrt { - 1} = i$ . the general form of a complex number is $a + ib$ here the $a$ is the real part and $ib$ is the imaginary part also in the above the roots of the given equation is $x = \dfrac{{ - 1 \pm i\sqrt 7 }}{{ - 2}}$ so here the real part is $\dfrac{{ - 1}}{{ - 2}}$ and the imaginary part is $\pm \dfrac{{i7}}{{ - 2}}$ .