Question

iIf the equation $y = rx{^2} + sx + t$ where r, s and t are real number such that $|s| < 1,|r| > |t| > 1,and$ $r.t > 0$ what is true about the zeros of the function y
A.There is one real zero of y
B.There are two real zeros of y
C.There is one complex zero of the function y
D.There are two complex zero of the function y
E.Not enough information

Answer 1

Hint: The nature of roots/zeros depends upon the nature of discriminant of the quadratic equation and complex roots always occur in pairs. if the discriminant is positive then roots are real, if the discriminant is negative then the roots are imaginary and zeros are complex in nature.

Complete step-by-step answer:
The equation $y = rx{^2} + sx + t$
$\eqalign{
  & |s| < 1,means \cr
  & - 1 < s < 1 \cr
  & s \in ( - 1,1) \cr} $
And
$\eqalign{
  & |r| > 1,means \cr
  & r \in ( - \infty , - 1) \cup (1,\infty ) \cr
  & |t| > 1,means \cr
  & t \in ( - \infty , - 1) \cup (1,\infty ) \cr
  & \cr} $
Now discriminant of a quadratic equation is
$${\sqrt {{{(coefficient{\text{ }}of{\text{ }}x)}^{}} - 4(coefficient{\text{ }}of{\text{ }}{x^2})(cons\tan t)} ^{}}$$
Here coefficient of $x = s$
And coefficient of ${x^2} = r$
Constant $t = t$
Therefore, after substituting the values in the formula of discriminant we get
$$discriminant = \sqrt {{s^2} - 4rt} $$
Now we know that $r.t > 1$
Multiplying both side of the equation with 4 we get
$4r.t > 4$
$$\eqalign{
  & s \in ( - 1,1) \cr
  & so{\text{ }}{s^2} \in [0,1) \cr
  & 0 < {s^2} < 1 \cr} $$
Subtracting $4r.t$ on both sides
$$\eqalign{
  & 0 - 4 < {s^2} - 4r.t < 1 - 4 \cr
  & - 4 < {s^2} - 4r.t < - 3 \cr} $$
 So the discriminant Is negative
 Let the discriminant is d
 $$d \in ( - \infty ,0)$$
Roots of the equation are $$roots = \dfrac{{ - s \pm \sqrt d }}{{2r}}$$
Since d is negative do square root of a negative number is complex
Which means there are complex roots of the equation and we have studied that complex roots always occur in pairs which means that the equation $y = rx{^2} + sx + t$ has both the complex roots
And roots are $\dfrac{{ - s + i\sqrt d }}{{2r}},\dfrac{{ - s - i\sqrt d }}{{2r}}$
So, the correct answer is “ $\dfrac{{ - s + i\sqrt d }}{{2r}},\dfrac{{ - s - i\sqrt d }}{{2r}}$”.

Note: Complex roots always occur in pairs whereas real roots may or may not occur in pairs and if the discriminant is equal to zero then the roots are of equal value and the equation comes out to be a perfect square of the given linear expression. In these types of questions, students are advised to transform the equation in general form and then find the discriminant.