Question

The zeroes of the quadratic polynomial ${x^2} + 99x + 127$ are
  (a) both positive
  (b) both negative
  (c) one positive and one negative
  (d) both equal

Answer 1

Hint – In this problem let one root of the given quadratic polynomial be $\alpha $ and the other one be $\beta $. Then use the relation between the sum of the roots and the product of the roots in terms of coefficients given in the quadratic polynomial. This will help formulation of two equations. Solve them to find the value of roots and choose the right answer.

Complete step-by-step solution -
Given quadratic equation is
${x^2} + 99x + 127$
Now as we know that zeroes of the quadratic equation are nothing but the roots of the quadratic equation.
Now, as we know that in a quadratic equation the sum of roots is the ratio of negative time’s coefficient of x to the coefficient of $x^2$.
Let the roots of this quadratic equation be $\alpha ,\beta $
$ \Rightarrow \alpha + \beta = \dfrac{{ - 99}}{1} = - 99$................ (1)
So as we see that the sum is negative so at least one of the roots is negative.
Now we also know that in a quadratic equation the product of roots is the ratio of constant term to the coefficient of $x_2$.
 $ \Rightarrow \alpha \beta = \dfrac{{127}}{1} = 127$.................... (2)
Now from equation (1) of the root is negative but from equation (2) the product of roots is positive this is only possible when both roots are negative (as multiplication of two negative numbers is positive number).
So both the roots of the given quadratic equation are negative.
So this is the required answer.
Hence option (A) is correct.

Note – There can be another method to solve this problem, in order to compute the roots we could have directly used Dharacharya formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for any quadratic equation of the form $a{x^2} + bx + c = 0$. This too gives the same answer.