how many real number solutions does this equation have? -7x^2+6x+3=0
How many real number solutions does the equation have? 0=3x^2+18x+27
a. B^2-4AC = 36-(-84) = 120.
The discriminant is positive. Therefore, we have 2 real solutions.
b. B^2-4AC = 324-324 = 0. 1 real solution.
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I apologize if my explanation was not clear. The solutions for a quadratic equation in the form of Ax^2 + Bx + C = 0 can be determined by the discriminant, which is the expression inside the square root of the quadratic formula. The discriminant is B^2 - 4AC.
If the discriminant is positive, then the quadratic equation has 2 real solutions.
If the discriminant is zero, then the quadratic equation has 1 real solution.
If the discriminant is negative, then the quadratic equation has 2 complex solutions.
Using these rules, we can solve for the number of real solutions in the given equations:
a. -7x^2 + 6x + 3 = 0
B^2 - 4AC = (6)^2 - 4(-7)(3) = 120
Since the discriminant is positive, we have 2 real solutions.
b. 0 = 3x^2 + 18x + 27
B^2 - 4AC = (18)^2 - 4(3)(27) = 0
Since the discriminant is zero, we have 1 real solution.
Well, let's find out! We can use the discriminant to determine the number of real number solutions. For the first equation, -7x^2+6x+3=0, the discriminant is b^2-4ac = 6^2 - 4(-7)(3) = 36 + 84 = 120. Since the discriminant is positive, we know that there are two real number solutions.
Now, for the second equation, 0=3x^2+18x+27, the discriminant is b^2-4ac = 18^2 - 4(3)(27) = 324 - 324 = 0. Since the discriminant is zero, we know that there is only one real number solution.
So, to summarize: the first equation has two real number solutions, and the second equation has one real number solution.
To determine the number of real number solutions for an equation, we can use the discriminant, which is a term in the quadratic formula.
For the equation -7x^2+6x+3=0, the discriminant is given by:
b^2 - 4ac
Where a, b, and c are the coefficients of the equation. In this case, a = -7, b = 6, and c = 3.
Substituting the values into the discriminant formula, we have:
6^2 - 4 * (-7) * 3
Simplifying further:
36 + 84
Which gives us:
120
Since the discriminant is positive (120 > 0), the equation has two real number solutions.
Now, let's move on to the second equation:
0=3x^2+18x+27
Again, we calculate the discriminant using the coefficients a = 3, b = 18, and c = 27:
b^2 - 4ac
18^2 - 4 * 3 * 27
324 - 324
The result is 0.
Since the discriminant is zero (0 = 0), the equation has only one real number solution.
In summary:
- The equation -7x^2+6x+3=0 has two real number solutions.
- The equation 0=3x^2+18x+27 has one real number solution.