Question

How do you find the discriminant and define the nature of the roots of the equation\[ - 4{m^2} - 4m - 5\]?

Answer 1

Hint: Here in this question you have to learn the well known equation given by siddhacharya known as siddhacharya equation which helps in finding the roots of quadratic equation and nature of roots of quadratic equation. 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: \[ - 4{m^2} - 4m - 5\]
Comparing this equation with general quadratic equation \[a{x^2} + bx + c\]
We get,
Value of \[a = - 4,b = - 4,c = - 5\]
Now, determinant D equals
\[
  D = \sqrt {{b^2} - 4ac} \\
   = \sqrt {{{( - 4)}^2} - 4( - 4)( - 5)} \\
   = \sqrt {16 - 80} \\
   = \sqrt {( - 64)} \\
   = \sqrt { - 1} \times \sqrt {{8^2}} \\
   = 8i\,(\sin ce\,\sqrt { - 1} = i) \\
 \]
Here we are getting the negative value of discriminant hence we will get complex roots.
Now, for roots of equation we use
\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
We get,
 \[
   = \dfrac{{ - ( - 4) + 8i}}{{2( - 4)}},\,\dfrac{{ - ( - 4) - 8i}}{{2( - 4)}} \\
   = \dfrac{{ - (8i + 4)}}{8},\,\dfrac{{ - ( - 8i + 4)}}{8} \\
   = - (2i + 1),\, - ( - 2i + 1) \\
   = (1 - 2i),( - 1 + 2i) \\
 \]
Hence roots of the equation are \[(1 - 2i),( - 1 + 2i)\] and the nature of roots are complex.
Formulae Used: Siddhacharya rule which says for any quadratic equation \[a{x^2} + bx + c\] the detrerminant 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.
And \[\sqrt { - 1} = i\]

Note:
Here in this question midterm 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 midterm splitting method also.