Question

After I used quadratic formula for -3x^2+6x-5 13 years ago

Answer 1

well I'd use the discriminant

\[b^2 - 4ac \]

because if \[b^2 - 4ac < 0 \]

and a is negative, then the curve is negative definite

so

36 - 4 x (-3)(-5) < 0

-24 < 0 so the curve has no real roots... and since a = -3 the curve is concave down.

the curve has an axis is symmetry at x = 1 and a vertex at (1, -2)

so the curve looks like



so

\[-3x^2 + 6x - 5 \le 0, ..(-\infty, \infty)\]

Answer 2

or the curve is <= 0 for all real x

Answer 3

wow...thanks! why did you use only b^2 - 4ac?

Answer 4

because the discriminant gives you information about the parabola

> 0 means real roots which can be rational or irrational

= 0 means equal or repeated roots.

< 0 no real roots.

so once you know there are no real roots... the curve is always positive ( positive definite) or in your case always negative ( negative definite)

actually my 1st thought was to graph the curve, when I saw you results for the GQF... I knew it was always negative... but just needed to show you

Answer 5

how did you get symmetry at x = 1 and a vertex at (1, -2) ?

Answer 6

ok... the line of symmetry for a parabola, y = f(x) is found using

\[x =\frac{-b}{2a}\]

which you may have seen in the GQF...

so

x = -6/(2x-3)

x = 1

Answer 7

so. much. detail. hehe. when finding the answer, would it be like x<or equal to something? i'm a bit confused.

Answer 8

ok.... lets look at a couple

x\[x^2 + 5x - 6 < 0\]

so

\[(x + 6)(x - 1) < 0\]

graphing it



so the question is asking when is the curve below the x axis

-6 < x< 1

is the algebraic solution.

in these questions you are only seeking solutions for real numbers... if you get a complex solution, then the curve doesn't cut the x- axis