State the value of the discriminant and the number of real solutions.
t^2+14t+49=0
The value of the discriminant is 0 and the equation has no real solution.
No
If the discriminant is zero, then the quadratic factors to a perfect square, and there will be one real solution (actually two real but equal solutions)
t^2+14t+49=0
(t+7)^2 = 0
t+7 = 0
t = -7
To find the value of the discriminant, we can use the formula:
Discriminant (D) = b^2 - 4ac
In the given quadratic equation, t^2 + 14t + 49 = 0, we can identify the values of a, b, and c:
a = 1 (coefficient of t^2 term)
b = 14 (coefficient of t term)
c = 49 (constant term)
Plugging these values into the discriminant formula, we get:
D = (14)^2 - 4(1)(49)
= 196 - 196
= 0
Now, let's determine the number of real solutions based on the value of the discriminant.
If the discriminant is positive (D > 0), the quadratic equation has two distinct real solutions.
If the discriminant is zero (D = 0), the quadratic equation has one real solution (also known as a repeated or double root).
If the discriminant is negative (D < 0), the quadratic equation has no real solutions; it only has imaginary solutions.
In this case, since the discriminant is 0, we can conclude that the equation t^2 + 14t + 49 = 0 has no real solutions.