When b^2 - 4ac > 0 --> 2 intercept
When b^2 - 4 ac = 0 --> 1 intercept
When b^2 - 4 ac <0> No intercept
(1) a>0
This does not tell us anything because b^2 - 4ac can be either negative or positive
E.g.:
- If b=1, a = 1, C = 4, 1^2 - 4 x 1 x 4 = 1 - 16 = -15
- If b=5, a = 1, C = 4, 5^2 - 4 x 1 x 4 = 25 - 16 = 9
(2) C=-6
This does not tell us anything because b^2 - 4ac can be either negative or positive
(1) + (2)
b^2: will 0 or positive
- 4ac: will positive
--> We will have 2 intercepts
Where did you get that question btw? Because it is very similar to one I had on the actual GMAT...
Help me about the concepts
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Source: Beat The GMAT — Problem Solving |
I am practising some math questions from various books (kaplan, PR, and OG). In one of the books I have found such type of question. BUt I am not clear about the concepts of Quadratic Equation.
The another question: What is the X intercept?
Actually, I wanted to know the concept of how to approach such type of questions.
Thanks in advance?
The another question: What is the X intercept?
Actually, I wanted to know the concept of how to approach such type of questions.
Thanks in advance?
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Magellan
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X intercept is the value of X such as Y = 0
For instance:
-) Y = 4 X, Y = 0 when X = 0
-) Y = 4X + 2, Y = 0 when X = -1/2
Whith an equation of the form y = ax^2 + bx + c, you should compute E = b^2 - 4ac
If E = b^2 - 4ac > 0, there are two intercepts given by (-b+sqrt(E))/2a and (-b-sqrt(E))/2a
If E = b^2 - 4ac = 0, there is one intercept given by -b/2a
If E = b^2 - 4ac < 0, there is no intercept
For instance:
-) Y = 4 X, Y = 0 when X = 0
-) Y = 4X + 2, Y = 0 when X = -1/2
Whith an equation of the form y = ax^2 + bx + c, you should compute E = b^2 - 4ac
If E = b^2 - 4ac > 0, there are two intercepts given by (-b+sqrt(E))/2a and (-b-sqrt(E))/2a
If E = b^2 - 4ac = 0, there is one intercept given by -b/2a
If E = b^2 - 4ac < 0, there is no intercept
