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does anyone have a way that might make a little more sence to me??? the way that i have.. is not doing it.... what are the possitive and negative signes above the numberline??? why are they there??? how do i figure this out????

2006-09-14 03:45:37 · 4 answers · asked by shenae 2 in Science & Mathematics Mathematics

4 answers

They serve a definite purpose. Think about a point that lies somewhere above the number line: Where exactly does the point lie? Well, to specify you must explain how far away the point is from the number line and on which side of the number line the point lies. If the point lies on the top side of the number line we say it is on the positive side; if the point is on the bottom side we say it is on the negative side.

2006-09-14 03:56:51 · answer #1 · answered by bruinfan 7 · 0 0

The + & - signs in the numerator of the quadratic eq mean there are 2 answers to the equation. One answer is found using the + sign, the other using the - sign. It's like writing out 2 complete equations that are different only in that single area (the sign). A sort of math shorthand.

Quadratics often have 2 answers because they are curves that cut through the x-axis in 2 places. They can also touch the x-axis at just one point, in which case the value of the radical following the +- is zero. If the graph of the function doesn't intersect the x-axis at all, the value of b^2-4ac is negative and the root of it is imaginary.

2006-09-14 11:04:13 · answer #2 · answered by Steve 7 · 0 0

consider ax^2+bx+c>0
let ax^2+bx+c=a(x-p)(x-q)
let p case 1 if x

so (x-p)(x-q)>0
case 2. if x>q then x-p>0 and x-q>0
so(x-p)(x-q)>0
so if (x-p)(x-q)>0 then the value of x lies outside p and q
consider ax^2+bx+c=0 <0
let ax^2+bx+c=a(x-p)(x-q)
let p then x-p>0 and x-q<0
so (x-p)(x-q)<0
that is if (x-p)(x-q)<0 then the values of x lie between pand q

2006-09-14 11:39:34 · answer #3 · answered by raj 7 · 0 0

do trial and error method.. it will definitely work...

2006-09-14 10:49:22 · answer #4 · answered by tyron01 2 · 0 1

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