Classify each quadratic equation as having either one or two real solutions.
One Real Solution
Two Real Solutions
X^2+5x=0
X^2−10x+25=0
X^2+10x+25=0
X^2-25=0
One Real Solution: X^2-25=0
Two Real Solutions: X^2+5x=0, X^2−10x+25=0, X^2+10x+25=0
To determine the number of real solutions for each quadratic equation, we can make use of the discriminant (denoted as Δ). The discriminant is calculated using the formula Δ = b^2 - 4ac, where a, b, and c are the coefficients in the quadratic equation ax^2 + bx + c = 0.
1. For the quadratic equation X^2+5x=0:
The equation is already in standard form (ax^2 + bx + c = 0), where a = 1, b = 5, and c = 0.
Using the discriminant, Δ = b^2 - 4ac = 5^2 - 4(1)(0) = 25.
Since Δ > 0, the quadratic equation X^2+5x=0 has two real solutions.
2. For the quadratic equation X^2−10x+25=0:
Again, the equation is already in standard form, where a = 1, b = -10, and c = 25.
Δ = b^2 - 4ac = (-10)^2 - 4(1)(25) = 100 - 100 = 0.
Since Δ = 0, the quadratic equation X^2−10x+25=0 has only one real solution.
3. For the quadratic equation X^2+10x+25=0:
Standard form: a = 1, b = 10, c = 25.
Δ = b^2 - 4ac = 10^2 - 4(1)(25) = 100 - 100 = 0.
Similar to the previous equation, Δ = 0, meaning it also has one real solution.
4. For the quadratic equation X^2-25=0:
Standard form: a = 1, b = 0, c = -25.
Δ = b^2 - 4ac = 0^2 - 4(1)(-25) = 0 + 100 = 100.
Since Δ > 0, the quadratic equation X^2-25=0 has two real solutions.
So, to summarize:
- X^2+5x=0 has two real solutions.
- X^2−10x+25=0 has one real solution.
- X^2+10x+25=0 has one real solution.
- X^2-25=0 has two real solutions.
To classify whether a quadratic equation has one or two real solutions, we can use the discriminant. The discriminant is the part of the quadratic formula inside the square root sign, which is equal to b^2 - 4ac for the quadratic equation ax^2 + bx + c = 0.
For each equation:
1. X^2 + 5x = 0:
From this equation, we can see that a = 1, b = 5, and c = 0. By substituting these values into the discriminant formula, we get:
Discriminant = 5^2 - 4(1)(0) = 25 - 0 = 25.
Since the discriminant (25) is positive, this equation has two real solutions.
2. X^2 - 10x + 25 = 0:
Here, a = 1, b = -10, and c = 25. By substituting these values into the discriminant formula, we get:
Discriminant = (-10)^2 - 4(1)(25) = 100 - 100 = 0.
Since the discriminant (0) is zero, this equation has one real solution.
3. X^2 + 10x + 25 = 0:
Again, a = 1, b = 10, and c = 25. Substituting these values into the discriminant formula gives us:
Discriminant = 10^2 - 4(1)(25) = 100 - 100 = 0.
Since the discriminant (0) is zero, this equation has one real solution.
4. X^2 - 25 = 0:
For this equation, a = 1, b = 0, and c = -25. Plugging these values into the discriminant formula gives us:
Discriminant = 0^2 - 4(1)(-25) = 0 + 100 = 100.
Since the discriminant (100) is positive, this equation has two real solutions.
Therefore, the classifications are as follows:
X^2 + 5x = 0: Two Real Solutions
X^2 - 10x + 25 = 0: One Real Solution
X^2 + 10x + 25 = 0: One Real Solution
X^2 - 25 = 0: Two Real Solutions