More than two solutions.
This equation is satisfied for all values of t lying between 2 and 3, i.e., 2 ≤ t≤ 3.
Thus, the given equation is satisfied for all values of x lying between 5 and 10.
If both the roots of the equation x2 – 6ax + 2 – 2a + 9a2 = 0 exceed 3, then
Let α, β be the roots of the equation x2 – ax + b = 0 and An = αn + βn.
Let f (x) = ax2 + bx + c, a, b, c Ïµ R. if f (x) takes real values for real values of x and non – real values for non – real values of x, then.
If α, β, γ are such that α + β + γ = 2, α2 + β2 + γ2 = 6, α3 + β3 + γ3 = 8, then α4 + β4 + γ4 is
The condition that the equation has real roots that are the equal in magnitude but opposite in sign is
If x is real, and
Let a > 0, b > 0 and c > 0. Then both the roots of the equation ax2 + bx +c = 0
Let a, b, c be non – zero real number such that
Then the quadratic equation ax2 + bx + c = 0 has
The equation 3x–1 + 5x–1 = 34 has