Which equation has no real solutions?
b^2 + 3b = - 3
2c^2 + 4c = 9
-5c^2 - c =-2
7g + 1 + 3g^2 = 0
the one with a negative discriminant
To determine which equation has no real solutions, we need to check the discriminant of each equation. The discriminant can be calculated using the quadratic formula, which is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
If the discriminant (b^2 - 4ac) is less than 0, the equation has no real solutions. Let's calculate the discriminant for each equation:
1. For the equation b^2 + 3b = -3:
a = 1, b = 3, c = -3
Discriminant = 3^2 - 4(1)(-3) = 9 + 12 = 21
Since the discriminant (21) is greater than 0, this equation has real solutions.
2. For the equation 2c^2 + 4c = 9:
a = 2, b = 4, c = -9
Discriminant = 4^2 - 4(2)(-9) = 16 + 72 = 88
Since the discriminant (88) is greater than 0, this equation has real solutions.
3. For the equation -5c^2 - c = -2:
a = -5, b = -1, c = -2
Discriminant = (-1)^2 - 4(-5)(-2) = 1 - 40 = -39
Since the discriminant (-39) is less than 0, this equation has no real solutions.
4. For the equation 7g + 1 + 3g^2 = 0:
a = 3, b = 7, c = 1
Discriminant = 7^2 - 4(3)(1) = 49 - 12 = 37
Since the discriminant (37) is greater than 0, this equation has real solutions.
Therefore, the equation -5c^2 - c = -2 has no real solutions.
To determine which equation has no real solutions, we need to use the concept of the discriminant in quadratic equations.
The general form of a quadratic equation is: ax^2 + bx + c = 0, where a, b, and c are constants.
For the given equations, let's analyze each one to find their discriminants:
1. Equation: b^2 + 3b = -3
Rearranging it, we have: b^2 + 3b + 3 = 0
The discriminant is: Δ = b^2 - 4ac
In this case, a = 1, b = 3, and c = 3.
So, Δ = 3^2 - 4(1)(3) = 9 - 12 = -3
Since the discriminant is negative, the equation has no real solutions.
2. Equation: 2c^2 + 4c = 9
Rearranging it, we have: 2c^2 + 4c - 9 = 0
The discriminant is: Δ = b^2 - 4ac
In this case, a = 2, b = 4, and c = -9.
So, Δ = 4^2 - 4(2)(-9) = 16 + 72 = 88
Since the discriminant is positive, the equation has two real solutions.
3. Equation: -5c^2 - c = -2
Rearranging it, we have: -5c^2 - c + 2 = 0
The discriminant is: Δ = b^2 - 4ac
In this case, a = -5, b = -1, and c = 2.
So, Δ = (-1)^2 - 4(-5)(2) = 1 + 40 = 41
Since the discriminant is positive, the equation has two real solutions.
4. Equation: 7g + 1 + 3g^2 = 0
Rearranging it, we have: 3g^2 + 7g + 1 = 0
The discriminant is: Δ = b^2 - 4ac
In this case, a = 3, b = 7, and c = 1.
So, Δ = 7^2 - 4(3)(1) = 49 - 12 = 37
Since the discriminant is positive, the equation has two real solutions.
Therefore, the equation "b^2 + 3b = -3" has no real solutions.