Question

Write the solution set of the equation $ {x^2} + x - 2 = 0 $ in roster form?

Answer 1

Hint: In order to determine the factors of the above quadratic equation use the Splitting up the middle term method or also called mid-term factorization method. In which the middle term would be splitted into two parts and similar values will be taken common and equated to zero to obtain the roots. Then place the roots in the form of sets, which is known as roster form.

Complete step-by-step answer:
Given a quadratic equation $ {x^2} + x - 2 = 0 $ ,let it be $ f(x) $
 $ f(x) = {x^2} + x - 2 $
Comparing the equation with the standard Quadratic equation $ a{x^2} + bx + c $
 $ a $ becomes $ 1 $
 $ b $ becomes $ 1 $
And $ c $ becomes $ - 2 $
To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of $ {x^2} $ and the constant term which comes to be
 $ = 1 \times - 2 = - 2 $
Now the second Step is to find the $ 2 $ factors of the number $ - 2 $ such that whether, addition or subtraction of those numbers is equal to the middle term or coefficient of $ x $ and the product of those factors results in the value of constant.
So, if we factorize $ - 2 $ , the answer comes to be $ 2 $ and $ - 1 $ as $ 2 - 1 = 1 $ that is the middle term. and $ 2 \times \left( { - 1} \right) = - 2 $ which is perfectly equal to the constant value.
Now writing the middle term sum of the factors obtained, so equation $ f(x) $ becomes
 $ f(x) = {x^2} + 2x - x - 2 $
Now taking common from the first $ 2 $ terms and last $ 2 $ terms
 $ f(x) = x(x + 2) - 1(x + 2) $
Finding the common binomial parenthesis, the equation becomes
 $ f(x) = (x - 1)(x + 2) $
Hence, we have successfully factorized our quadratic equation.
Therefore, the factors are $ \left( {x - 1} \right) $ and $ \left( {x + 2} \right) $ .
Equating the factors with zero to get the roots:
 $
\Rightarrow \left( {x - 1} \right) = 0 \\
\Rightarrow x = 1 \;
 $
And,
 $
\Rightarrow \left( {x + 2} \right) = 0 \\
\Rightarrow x = - 2 \;
 $
Therefore, the roots of the quadratic equation $ f(x) = {x^2} + x - 2 $ are: $ x = 1 $ and $ x = - 2 $ .
Since, we know that in roster form the elements are listed in sets, separated by commas, inside the curly brackets.
Hence, the roster form for the equation $ {x^2} + x - 2 = 0 $ is $ \left\{ {1, - 2} \right\} $ .
So, the correct answer is “ $ \left\{ {1, - 2} \right\} $ ”.

Note: You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
 $ {x_1} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} $ and $ {x_2} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} $
 $ {x_1} $ , $ {x_2} $ are root to quadratic equation $ a{x^2} + bx + c $
 Hence the factors will be $ \left( {x - {x_1}} \right)and\,\left( {x - {x_2}} \right) $ .
Then write the values in roster form.