Solve f(x) > 0

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sirhc

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Hello,

I've been stuck on this final question for a while. I've completed the first questions in the problem correctly (well hopefully).

1. Let f(x) = 2x^2 − 8x − 7.

(a) Write the quadratic in vertex form.
= 2(x-2)^2-1

(b) Calculate the roots of the function f.
x = -0.739
x = 4.739

(c) Sketch the graph of f, showing all important features. (Vertex, roots, y-intercept.)
Vertex (2,-15)
Roots -0.739 and 4.739
y-intercept -7

(d) Solve f(x) > 0

the only idea I had was;

_____________
8x (+-) / 8x-4(2x^2)-7
------------------------------
2(2x^2)


Any help with this would be greatly appreciated.

Thanks,
sirhc :smile:
 
Last edited:
sirhc said:
f(x) = 2x^2 − 8x − 7.

(a) Write the quadratic in vertex form.

2(x - 2)^2 - 39 This value is not correct

After you fix it, you will need to redo parts (b) and (c).

(d) Solve f(x) > 0
Click to expand...
You know that symbol f(x) is another way to write y.

Hint: In terms of a parabolic graph, what is the meaning to say that y > 0?

If you need more help, please show us what you thought about or tried. :cool:

PS: It's not necessary to report decimal answers to the nearth ten-billionth, unless you've been instructed to do that. If you report a number to three decimal places, be sure to properly round it.
 
Thanks for the reply. I've thought about it and come up with a few ideas;


Distance between the roots. 4.739 - 0.739 = 4

or

y = (x + 0.739) (x - 4.739)
x < 0.739 or x > 4.739

Cheers,
sirhc
 
sirhc said:
f(x) = 2x^2 − 8x − 7.

Write the quadratic in vertex form.

2(x - 2)^2 - 1

This value is not correct. Please show your work on completing the square.

(b) Calculate the roots of the function f.

x = -0.739

x = 4.739

(c) Sketch the graph of f, showing all important features. (Vertex, roots, y-intercept.)

Vertex (2,-15)

Roots -0.739 and 4.739

y-intercept -7
Click to expand...
The rest of the information above looks correct.

The exact roots are:

x = 2 + sqrt(30)/2

x = 2 - sqrt(30)/2

:cool:
 
f(x) = 2x^2 - 8x - 7

y=2(x^2 - 4x) - 7
y=2(x^2 - 4x - 4) - 7 + 8

= 2(x - 2)^2 - 1

-----------------------------

Distance between the roots. 4.739 -- 0.739 = 5.478

y = (x + 0.739) (x - 4.739)
x < - 0.739 or x > 4.739

^so is the answer to solve f(x) > 0
 
sirhc said:
f(x) = 2x^2 - 8x - 7

y=2(x^2 - 4x) - 7

y=2(x^2 - 4x - 4) - 7 + 8 Thank you for showing your work.

Those two arithmetic operators are not correct.

Can you see why? :cool:

-----------------------------

Distance between the roots. 4.739 -- 0.739 = 5.478

y = (x + 0.739) (x - 4.739)

x < - 0.739 or x > 4.739

^so is the answer to solve f(x) > 0
Click to expand...
Yes. That is the correct solution to the given inequality (reported as decimal approximations).
 
oh...

y=2(x^2 - 4x - 4) - 7 + 8


= 2(x - 2)^2 + 1


Thanks so much,
sirhc
 
sirhc said:
oh...

y=2(x^2 - 4x - 4) - 7 + 8
Click to expand...
Yes, that is what I posted.

I asked whether you can see why the parts highlighted in red are wrong.


2(x - 2)^2 + 1
Click to expand...

I also told you already that this is not correct.

Look what happens, if we multiply out this result and combine like terms:

2(x^2 - 4x + 4) + 1

2x^2 - 8x + 8 + 1

2x^2 - 8x + 9

We do not get the original function f.

Can you find your mistake? :cool:
 
sirhc said:
y=22(x^2-4x+4)-7+8

That's a typo

the minus instead of the positive

Yes, you wrongly added -4, to complete the square, instead of adding 4.

(b/2)^2 = (-4/2)^2 = (-2)^2 = 4
Click to expand...

Now, let's fix the + 8.

Because of the multiplication by 2, your added 4 results in a total of adding 8, so you need to subtract that 8, outside of the parentheses, right?

So, what is your final answer, for completing the square?
 
= 2(x^2 - 4x + 4) - 15
= (2x^2 - 8x + 8) - 15
= (2x^2 - 8x) - 15 + 8

= 2x^2 − 8x − 7
 
sirhc said:
2(x^2 - 4x + 4) - 15
Click to expand...
That's not how I would report the answer for completing the square.

The correct form is: f(x) = 2(x-2)^2 - 15

I am glad to see that you understand the basic idea of completing the square. The various mistakes that you have made in this exercise concern signs. With greater care and more practice, you will make less sign errors.

Cheers :cool:
 
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