College Algebra Dummie I need HELP!!!!?

Question

x^2-6x+4>0 I know the multiplying factor is 4 and the adding factor is -6 but I don't know where to go from there.

9x^2-6x-4<0 I know the multiplying factor is -36 and adding factor is -6 where would I go from there b/c nothing I seem to try will go with those numbers. 8*(

Please Help!!!

Answer 1

Neither of those will factor. You need to use the quadratic formula:

if Ax^2 + Bx + C = 0,
then x = -b/2a +/- sqrt(b^2-4ac)/2a.

For x^2 - 6x + 4, A = 1, B = -6, and C= 4.

So x^2 - 6x + 4 = 0 when
x = -(-6)/2(1) +/- sqrt( (-6)^2 - 4(1)(4) )/2(1)
x = 3 +/- sqrt( 20 )/2
x = 3 +/- sqrt(5)

Since the leading coefficient of x^2 - 6x + 4 (number multiplying x^2) is positive (x^2 = 1x^2), the graph of y = x^2 - 6x + 4 is an upward opening parabola, and so y>0 when x is not between the roots from the quadratic formula. In other words,

x < 3-sqrt(5) or x > 3+sqrt(5)

or in interval notation, the solution to x^2 - 6x + 4 > 0 is

( -infinity, 3-sqrt(5) ] U [ 3+sqrt(5), +infinity )

Answer 2

Let's try to solve these inequalities.
Here's a principle to guide you:
These curves can only change sign at a zero
of themselves.
So the first step is to find the roots
of f(x) =x² -6x+4 = 0.
They are
x = 3 + √5 = 5.536(approx)
and x = 3 -√5 = .7639(approx)
But f(1) = 5-6 = -1.
So the principle tells us that f(x) <0 between
its 2 roots and f(x)>0 everywhere else.
So the answer to the first problem
is x < 3-√5 and x > 3+√5.

Let's use the same ideas to do the second one:
Let f(x) =9x² -6x -4.
The roots are
1/18(6 + √180) = 1.079(approx)
and 1/18(6 - √180) = -0.412(approx)
But f(0) = -4 so f(x) <0 only between its 2 roots.
No, you are not a "dummie". You just needed some
guidance in doing this sort of problem!

Answer 3

1. you should use the discriminant D=b^2 -4ac, x = (-b +/- sqrt D)/2a

Answer 4

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