Solution:
The given equation is: t2x2 x 9 = 0
Let’s find the roots of the given equation using the quadratic formula
Quadratic formula: For a quadratic equation ax2+bx+c=0, the roots are given by the formula:
x = {-b±√(b2-4ac)}/2a
Given equation is: t2x2 x 9 = 0
Comparing with the standard quadratic equation ax2+bx+c=0, we get
a = t2, b = 1, and c = 9
Substituting the values of a, b, and c in the quadratic formula, we get
x = {-1±√(1-4t2×9)}/(2t2)
= {-1±√(1-36t2)}/(2t2)
The roots of the given equation are real when the discriminant is greater than or equal to 0.
Discriminant = b2-4ac
= 1-4t2×9
= 1-36t2
For the product of the roots to exist, both the roots should be real. This means the discriminant should be greater than or equal to 0.
1-36t2 ≥ 0
t2 ≤ 1/36
t ≤ 1/6 or t ≥ -1/6
If t = 0, then the given equation becomes 9 = 0 which is not possible. Hence, t cannot be equal to 0.
Therefore, the possible values of t are t < -1/6="" or="" t="" /> 1/6.
For both these values of t, the discriminant is greater than or equal to 0. Hence, both the roots are real.
Product of the roots = (first root) × (second root)
= {-1+√(1-36t2)}/(2t2) × {-1-√(1-36t2)}/(2t2)
= (1-36t2)/(4t4)
As t approaches either -∞ or ∞, the product of the roots approaches 0.
However, for any finite value of t that satisfies t < -1/6="" or="" t="" /> 1/6, the product of the roots does not exist since the denominator becomes 0.
Hence, the correct option is (a) does not exist.