Asked by HELP PLEASE
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
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
Answers
Answered by
Reiny
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:
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:
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