
Solve the equation $ {x^2} - x - 2 = 0 $
Answer
498.3k+ views
Hint: The given equation has the highest degree of two which means it is a quadratic equation. We can solve quadratic equations by two methods i.e. 1) Factoring method 2) Standard formula. We can use either of two to solve the given problem and find the values of the variable which is known as ‘roots of the equation’.
Complete step-by-step answer:
Firstly, we write down the equation $ {x^2} - x - 2 = 0 $
Now we see that it is an equation with degree two (the highest power of the given polynomial equation is two). It also has the same form as the general form of quadratic equation i.e.
$ a{x^2} + bx + c = 0\;;\;a,b\;\& \;c\; \in \mathbb{R}\;;\;a \ne 0 $
Comparing the given equation with the standard equation gives us
$
a = 1 \\
b = - 1 \\
c = - 2 \;
$
Now we have to apply the standard formula to solve the equation and find the value of the variable.
The formula for finding the roots of a quadratic equation is given by
$ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Applying the quadratic formula, we get
$
x = \;\dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4 \times 1 \times ( - 2)} }}{{2 \times ( 1)}} = \dfrac{{1 \pm \;\sqrt {1 + 8} }}{{ 2}} \\
x\; = \dfrac{{1 \pm \sqrt 9 }}{{ 2}} \\
$
Taking (+) and (-) one by one, and putting the value of
$ \sqrt 9 = 3 $
We get
$ x = \dfrac{{1 + 3}}{{ 2}} = \dfrac{4}{{ 2}} = 2 $
And we get
$ x = \dfrac{{1 - 3}}{{ 2}} = \dfrac{{ - 2}}{{ 2}} = -1 $
This means we get two values (two roots) of the equation such that
$ x = \; - 2\;{\text{and}}\;1 $
So, the correct answer is “ $ x = \; 2\;{\text{and}}\;-1 $ ”.
Note: Both the methods which are used to find the solution of quadratic equation are equally relevant and appropriate. For complex equations we generally use the standard formula for solving the quadratic equation and for simpler equations we generally use factorizing method. In our case we have solved the problem by using quadratic formula.
Complete step-by-step answer:
Firstly, we write down the equation $ {x^2} - x - 2 = 0 $
Now we see that it is an equation with degree two (the highest power of the given polynomial equation is two). It also has the same form as the general form of quadratic equation i.e.
$ a{x^2} + bx + c = 0\;;\;a,b\;\& \;c\; \in \mathbb{R}\;;\;a \ne 0 $
Comparing the given equation with the standard equation gives us
$
a = 1 \\
b = - 1 \\
c = - 2 \;
$
Now we have to apply the standard formula to solve the equation and find the value of the variable.
The formula for finding the roots of a quadratic equation is given by
$ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Applying the quadratic formula, we get
$
x = \;\dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4 \times 1 \times ( - 2)} }}{{2 \times ( 1)}} = \dfrac{{1 \pm \;\sqrt {1 + 8} }}{{ 2}} \\
x\; = \dfrac{{1 \pm \sqrt 9 }}{{ 2}} \\
$
Taking (+) and (-) one by one, and putting the value of
$ \sqrt 9 = 3 $
We get
$ x = \dfrac{{1 + 3}}{{ 2}} = \dfrac{4}{{ 2}} = 2 $
And we get
$ x = \dfrac{{1 - 3}}{{ 2}} = \dfrac{{ - 2}}{{ 2}} = -1 $
This means we get two values (two roots) of the equation such that
$ x = \; - 2\;{\text{and}}\;1 $
So, the correct answer is “ $ x = \; 2\;{\text{and}}\;-1 $ ”.
Note: Both the methods which are used to find the solution of quadratic equation are equally relevant and appropriate. For complex equations we generally use the standard formula for solving the quadratic equation and for simpler equations we generally use factorizing method. In our case we have solved the problem by using quadratic formula.
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