Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
s^2√5 + s + √5 = 0
I'm confused because of the radical...
I would have written it as
√5 s^2 + s = √5 = 0 , so it wouldn't be confused with s^(2√5) + s +so it wouldn't be confused with s^(2√5)5 = 0
That would be a mess
a = √5, b=1 , c=√5
b^2 - 4ac
= 5 - 4(1)√5
= 5 - 4√5
To find the discriminant of the quadratic equation, we first need to rewrite the equation in the standard form: ax^2 + bx + c = 0.
Let's do that with the given equation, s^2√5 + s + √5 = 0:
Start by moving √5 to the other side of the equation:
s^2√5 + s = -√5
Next, rewrite s^2√5 as √5 * s^2:
√5 * s^2 + s = -√5
Now, let's compare this equation to the standard form ax^2 + bx + c = 0:
a = √5
b = 1
c = -√5
The discriminant (denoted as Δ) is calculated as Δ = b^2 - 4ac.
Substituting the values into the formula, we get:
Δ = (1)^2 - 4(√5)(-√5)
= 1 - 4(5)
= 1 - 20
= -19
The discriminant, Δ, is -19.
Now, let's determine the number and type of solutions based on the value of the discriminant:
1. If Δ > 0, then there are two distinct real solutions.
2. If Δ = 0, then there is one real solution.
3. If Δ < 0, then there are no real solutions.
In this case, since Δ = -19, which is less than 0, there are no real solutions.
Therefore, the quadratic equation s^2√5 + s + √5 = 0 has no real solutions.