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.
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
3 answers
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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.
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.
1 answer