2x^2+3x-3=0

x=(-3 plus/minus $\sqrt {33}$)/4

x=-2.185 and x=0.685

Is this correct? Is there a better way to solve it?

Thank you

2. Originally Posted by oceanmd
2x^2+3x-3=0

x=(-3 plus/minus $\sqrt {33}$)/4

x=-2.185 and x=0.685

Is this correct? Is there a better way to solve it?

Thank you
it is x = -2.186 and x = 0.686

3. ## Slope of Secant Line

How to find the slope of the secant line for f(x) =2x^2-3x. Please solve it, as I have no idea how to do it.

Thank you

4. Originally Posted by oceanmd
How to find the slope of the secant line for f(x) =2x^2-3x. Please solve it, as I have no idea how to do it.

Thank you
we need an interval here. the slope of the secant line between which points?

5. I am doing the chapter review: a printout given by the teacher. It does not say between what points. Can you come up with the points and explain to me how to solve it.

Thank you

6. Originally Posted by oceanmd
I am doing the chapter review: a printout given by the teacher. It does not say between what points. Can you come up with the points and explain to me how to solve it.

Thank you
ok, the slope of the secant line of a function $f(x)$ between the points $x = a$ and $x = b$ is given by:

$\frac {f(b) - f(a)}{b - a}$

(this is just the formula for the slope of the straight line connecting the two points. note its similarity to $\frac {y_2 - y_1}{x_2 - x_1}$)

7. Ok, I will try
The two points A(0.75, -9/8) which is the vertex of the parabola
B (1.5,0), which is the x-intercept.
The slope of the line is 3/2

Is this correct?

8. Originally Posted by oceanmd
Ok, I will try
The two points A(0.75, -9/8) which is the vertex of the parabola
B (1.5,0), which is the x-intercept.
The slope of the line is 3/2

Is this correct?
yes