Question

How to factor \[2{x^2} + x - 1\] ?

Answer 1

Hint: We are given with a quadratic equation of the form \[a{x^2} + bx + c = 0\]. In order to find its root we will use the quadratic formula method here. We will get two roots of this equation since the degree of the equation is 2. Now we will use quadratic formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .Comparing the given equation with the general equation we will get the values of a, b and c. then putting them in formula we will find the answer!

Complete step-by-step answer:
Given the equation is \[2{x^2} + x - 1\] .
Generally we write the quadratic equation as \[a{x^2} + bx + c = 0\].
Now after comparing them we conclude that a=2, b=1 and c=-1.
Substituting them in quadratic formula we get,
\[ \Rightarrow \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 2 \times \left( { - 1} \right)} }}{{2 \times 2}}\]
Now solve the root first,
\[ = \dfrac{{ - 1 \pm \sqrt {1 + 8} }}{{2 \times 2}}\]
Now multiply the numbers in denominator and add numbers in root,
\[ = \dfrac{{ - 1 \pm \sqrt 9 }}{4}\]
We know that 9 is the perfect square of 3 so, \[\sqrt 9 = 3\]
\[ = \dfrac{{ - 1 \pm 3}}{4}\]
Now here onwards we will get two values of the expression,
\[ = \dfrac{{ - 1 + 3}}{4},\dfrac{{ - 1 - 3}}{4}\]
On solving this we get,
\[ = \dfrac{2}{4},\dfrac{{ - 4}}{4}\]
On simplifying these dfractions to simplest form,
\[ = \dfrac{1}{2}, - 1\]
And these are our roots.
\[2{x^2} + x - 1\] can be factored as \[x = \dfrac{1}{2}\] or \[x = - 1\].
So, the correct answer is “\[x = \dfrac{1}{2}\] or \[x = - 1\]”.

Note: Quadratic equations can be solved with the help of many methods. Like one we used above in which we used a quadratic formula and then we have a factorization method in which we factorize the middle term of the equation. Also the method of completing the square is also used .Note that the roots are sometimes equal when the discriminant \[{b^2} - 4ac\] is equal to 0. This discriminant part decides the nature of the roots.