Question

How do you solve ${x^2} - x - 20 = 0?$

Answer 1

Hint: The given equation is a quadratic equation that will get two solutions for $x$. Firstly find the discriminant of the equation and then try to solve it further.

Complete answer:
Given equation ${x^2} - x - 20 = 0$. For a quadratic equation $a{x^2} + bx + c = 0$, discriminant ($D$) is given by. $D = {b^2} - 4ac$
In the given equation we can see that the value of $a = 1$ , the value of $b = - 1$ and the value of $c = - 20$ so now putting the values of $a,\;b$ and $c$ in above equation to find the value of discriminant.
$
   \Rightarrow D = {b^2} - 4ac \\
   \Rightarrow D = {( - 1)^2} - 4 \times 1 \times ( - 20) \\
   \Rightarrow D = 1 + 80 \\
   \Rightarrow D = 81 \\
 $
Now, we get the value of discriminant which is greater than $0$, that means the roots of the equation ${x^2} - x - 20 = 0$ will be real and distinct. Read the note section of this question for more information.
To the given equation, we know that for a quadratic equation, roots are given by
$x = \dfrac{{1 + 9}}{2}$ _______(I)
So, we have the required values of $a,\;b$ and $D$ that will be substituted in equation (I) to find the solution or roots of the given equation.
Let us substitute the of $a,\;b$ and $D$ in equation (I), to get the required roots,
$
   \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\
   \Rightarrow x = \dfrac{{ - ( - 1) \pm \sqrt {81} }}{{2 \times 1}} \\
   \Rightarrow x = \dfrac{{1 \pm \sqrt {81} }}{2} \\
 $
As we know, $\sqrt {81} = \pm 9$, again solving further, we get
 $
   \Rightarrow x = \dfrac{{1 \pm ( \pm 9)}}{2} \\
   \Rightarrow x = \dfrac{{1 \mp 9}}{2} \\
 $
Here we got two solutions for $x$, simplifying further to get the roots,
 $ \Rightarrow x = \dfrac{{1 - 9}}{2}$ and $x = \dfrac{{1 + 9}}{2}$
$ \Rightarrow x = \dfrac{{ - 8}}{2}$ and $x = \dfrac{{10}}{2}$
$ \Rightarrow x = - 4$ and $x = 5$
Finally we got the required roots of the equation, which are$x = - 4$ and $x = 5$

Note: If you have the value of discriminant then you can easily tell about the nature of roots as follows
1. Roots will be real and distinct, if $D > 0$
2. Roots will be real and equal, if $D = 0$
3. Roots will be imaginary or no real roots, if $D < 0$
Where $D$ is the discriminant of the quadratic equation.