SOLUTION: True or False? Explain your answer briefly. (a) For any real number c , the quadratic equation x^2 + x - c^2 = 0 has two distinct (real) solutions. (b) If a > 4 , then the equati

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: True or False? Explain your answer briefly. (a) For any real number c , the quadratic equation x^2 + x - c^2 = 0 has two distinct (real) solutions. (b) If a > 4 , then the equati      Log On


   

Question 1122567: True or False? Explain your answer briefly.
(a) For any real number c , the quadratic equation x^2 + x - c^2 = 0 has two distinct (real) solutions.
(b) If a > 4 , then the equation ax^2 + 4x + 1 = 0 has no (real) solutions.
(c) If b^2 − 4ac ≥ 0, then the quadratic equation ax^2 + bx + c = 0 has at most one solution.
Help with this question would be greatly appreciated!
Found 4 solutions by greenestamps, josmiceli, ikleyn, Alan3354:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


For the quadratic equation ax%5E2%2Bbx%2Bc+=+0

the discriminant is b%5E2-4ac

It is called the DISCRIMINant because it DISCRIMINates between three cases:

(1) If the discriminant is positive, there are 2 real solutions
(2) If the discriminant is zero, there is exactly 1 real solution
(3) If the discriminant is negative, there are no real solutions

(a) The discriminant is 1%5E2-4%281%29%28-c%5E2%29+=+1%2B4c%5E2. What kinds of values (positive, zero, or negative) can that expression have?

(b) The discriminant is 4%5E2-4%28a%29%281%29+=+16-4a. Given that a > 4, what kinds of values can that expression have?

(c) The answer follows directly from the definition of the discriminant.

------------------------------------------------------------------
Reply to the student's question....

In the standard form of the quadratic equation, ax%5E2%2Bbx%2Bc=0, the "c" is whatever the constant term is.

In part (a) of your question, the constant term is "-c^2". That "c" is different than the "c" in the standard form of the quadratic equation. It just happens in this case that the constant term contains the variable "c".

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(a)
+x%5E2+%2B+x+-+c%5E2+=+0+
+%28+x+%2B+c+%29%2A%28+x+-+c+%29+=+0+
+x+=+-c+
+x+=+c+
2 real solutions
—————————-
(b)
The discrimination is
+b%5E2+-+4a%2Ac+
If +a+%3E+4+, then
+4%2Aa%2A1+%3E+16+
And
+b+=+4+
+b%5E2+=+16+
So, the discrimination is negative
and the solutions are imaginary ( not real )
——————————
(c)
If +b%5E2+-+4%2Aa%2Ac+=+0+, then there is 1 real solution
which is +x+=+-b%2F%282a%29+
But if +b%5E2+-+4%2Aa%2Ac+%3E+0+ there are 2 real solutions

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
Be aware :   The solution by  @josmiceli  related to the part  a)  -  is  INCORRECT  and  IRRELEVANT.



Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
All quadratics have 2 solutions.
If the discriminant = 0, then the 2 solutions are equal.
Likewise, all cubics have 3 solutions, quartics have 4, etc.