Which equation has exactly one real solution?
A 4x^2– 12x– 9 = 0
B 4x^2+ 12x+ 9 = 0
C 4x^2– 6x– 9 = 0
D 4x2+ 6x+ 9 = 0
Please teach me how to do this
Have you learned about b^2 - 4ac, called the discriminant, of the quadratic equation?
If the discriminant is zero, there is only one solution, so evaluate b^2 - 4ac.
If the discriminant is positive, there are 2 real and different solutions,
If the discriminant is negative, there are no real solutions
I will do A, you do the others to decide.
A:
4x^2 - 12x - 9 = 0 , a=4, b=-12, c = -9
b^2 -4ac
= 144-4(4)(-9) = + , so there are two different real solutions.
B:
To determine which equation has exactly one real solution, we can use the discriminant. The discriminant is a part of the quadratic formula and helps us identify the number of real solutions for a quadratic equation.
The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a
For a quadratic equation ax^2 + bx + c = 0:
1. If the discriminant (b^2 - 4ac) is positive, there are two distinct real solutions.
2. If the discriminant is zero, there is exactly one real solution.
3. If the discriminant is negative, there are no real solutions.
Now, let's apply this to each equation mentioned:
A) 4x^2 – 12x – 9 = 0
Here, a = 4, b = -12, and c = -9. Calculating the discriminant: b^2 - 4ac = (-12)^2 - 4(4)(-9) = 144 + 144 = 288. Since the discriminant is positive, this equation has two distinct real solutions.
B) 4x^2 + 12x + 9 = 0
Using the same process, the discriminant for this equation is: b^2 - 4ac = (12)^2 - 4(4)(9) = 144 - 144 = 0. As the discriminant is zero, this equation has exactly one real solution.
C) 4x^2 – 6x – 9 = 0
The discriminant for this equation is: b^2 - 4ac = (-6)^2 - 4(4)(-9) = 36 + 144 = 180. Since the discriminant is positive, this equation has two distinct real solutions.
D) 4x^2 + 6x + 9 = 0
The discriminant for this equation is: b^2 - 4ac = (6)^2 - 4(4)(9) = 36 - 144 = -108. Since the discriminant is negative, this equation has no real solutions.
Therefore, using the information from the discriminant, the equation with exactly one real solution is option B: 4x^2 + 12x + 9 = 0.
To determine which equation has exactly one real solution, we need to examine the discriminant of each quadratic equation. The discriminant is given by the formula Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
If the discriminant is greater than zero (Δ > 0), then the equation has two real solutions.
If the discriminant is equal to zero (Δ = 0), then the equation has exactly one real solution.
If the discriminant is less than zero (Δ < 0), then the equation has no real solutions.
Let's evaluate the discriminant for each equation:
A) 4x^2 - 12x - 9 = 0:
a = 4, b = -12, c = -9
Δ = (-12)^2 - 4(4)(-9) = 144 + 144 = 288
Since Δ > 0, this equation has two real solutions, not exactly one real solution.
B) 4x^2 + 12x + 9 = 0:
a = 4, b = 12, c = 9
Δ = (12)^2 - 4(4)(9) = 144 - 144 = 0
Since Δ = 0, this equation has exactly one real solution.
C) 4x^2 - 6x - 9 = 0:
a = 4, b = -6, c = -9
Δ = (-6)^2 - 4(4)(-9) = 36 + 144 = 180
Since Δ > 0, this equation has two real solutions, not exactly one real solution.
D) 4x^2 + 6x + 9 = 0:
a = 4, b = 6, c = 9
Δ = (6)^2 - 4(4)(9) = 36 - 144 = -108
Since Δ < 0, this equation has no real solutions.
Therefore, the equation B) 4x^2 + 12x + 9 = 0 has exactly one real solution.