Question

How do you solve the equation and find the factors by using quadratic formulae for \[{x^2} - 2x - 3 = 0\]

Answer 1

Hint: The rule says for any quadratic equation \[a{x^2} + bx + c\] the determinant of equation is \[\sqrt {{b^2} - 4ac} \] and roots can be obtained from \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , and nature of roots can be seen as real root or complex root.

Complete step-by-step answer:
Given equation: \[{x^2} - 2x - 3\]
Comparing this equation with general quadratic equation \[a{x^2} + bx + c\]
We get,
Value of \[a = 1,b = - 2,c = - 3\]
Now, determinant D equals
 \[
   \Rightarrow D = \sqrt {{b^2} - 4ac} \\
   \Rightarrow D = \sqrt {{{( - 2)}^2} - 4(1)( - 3)} \\
   \Rightarrow D = \sqrt {4 + 12} \\
   \Rightarrow D = \sqrt {16} \\
   \Rightarrow D = 4 \;
 \]
Here we are getting the negative value of discriminant hence we will get complex roots.
Now, for roots of equation we use
 \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Putting the values in the above equation, to get the roots of equation we get,
  \[
   \Rightarrow \dfrac{{ - ( - 2) + 4}}{{2(1)}},\,\dfrac{{ - ( - 2) - 4}}{{2(1)}} \\
   \Rightarrow \dfrac{6}{2},\,\dfrac{{ - 2}}{2} \\
   \Rightarrow 3, - 1 \;
 \]
Hence roots of the equation are \[3, - 1\] and the nature of roots are real.
So, the correct answer is “ \[3, - 1\] ”.

Note: Here in this question mid term splitting is not used for finding the roots because it is very difficult to think of the split terms for the middle term of the equation, which follows the condition of the mid term split rule for factorization. But if you think you can find such terms then definitely you can use a mid term splitting method also.