View Full Version : How is the distance to the x-axis related to the roots of quadratic equations?
lmamaths
Apr22-05, 07:34 AM
Hi,
How is the roots of a quadratic equation related
to the distance from the x-axis at where
the root is -
where ...
ax^2+bx+c=0
and ...
x = (-b +- SQRT(b^2-4ac))/2
Can someone help me to establish where this
distance relationship to the x-axis and the root
come from?
Thx!
LMA
HallsofIvy
Apr22-05, 09:15 AM
Am I understanding this correctly? Are you asking how the roots of a quadratic equation relate to "the distance from the x-axis at where the root is"? A root of an equation, by definition, is a point where y= 0. if y=0, then the distance from the x-axis is 0: the graph crosses the axis there!
lmamaths
Apr22-05, 10:58 AM
Hi,
Maybe I mean't lowest part of the curve to the x-axis, consider:
y=2x^2-3x+2
Thx!
Leo
Hi,
Maybe I mean't lowest part of the curve to the x-axis, consider:
y=2x^2-3x+2
Thx!
Leo
I'm not really sure what you're asking Leo. That function has no real roots. If you extend the domain to complex numbers then the function is still zero (both real and imaginary parts) at each of its complex zeros. Real or complex the function is still zero at it's zero's.
HallsofIvy
Apr22-05, 11:57 AM
Hi,
Maybe I mean't lowest part of the curve to the x-axis, consider:
y=2x^2-3x+2
Thx!
Leo
Ah- distance from the vertex to the x-axis.
Given y= ax2+ bx+ c= 0, consider solving by completing the square: write this as a(x2+ (b/a)x+ b2/4a[sup]2[sup])+ c- 4b2/a= a(x- (b/2a))2+ c- 4b2/a.
The distance from the x-axis to the vertex is given when x = b/2a and is
c- 4b2/a. do you see how that is connected to the value of x that satisfies the equation?
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