# Range and values of x

• May 9th 2013, 04:53 PM
Range and values of x
Hi guys, this is part of the harder portion of my homework, and I'm struggling to know where to start. Could someone point me in the right direction? Thanks a lot.

a) In this exercise, we find the range (i.e. the possible values) of (6x + 5) / (3x2 + 4x + 2)

Let D = (6x + 5) / (3x2 + 4x + 2) for all x.

i) Show that 3Dx2 + (4D - 6)x + 2D - 5 = 0
ii) Hence, show that 3Dx2 + (4D - 6)x + 2D - 5 = 0 has real solutions in x if and only if (D - 3)(2D + 3) <= 0
iii) Hence, find the range of (6x + 5) / (3x2 + 4x + 2)

b) Find the range of (x2 - 6x + 5) / (x2 + 2x + 1)
• May 9th 2013, 05:39 PM
Gusbob
Re: Range and values of x
Quote:

Hi guys, this is part of the harder portion of my homework, and I'm struggling to know where to start. Could someone point me in the right direction? Thanks a lot.

a) In this exercise, we find the range (i.e. the possible values) of (6x + 5) / (3x2 + 4x + 2)

Let D = (6x + 5) / (3x2 + 4x + 2) for all x.

i) Show that 3Dx2 + (4D - 6)x + 2D - 5 = 0
ii) Hence, show that 3Dx2 + (4D - 6)x + 2D - 5 = 0 has real solutions in x if and only if (D - 3)(2D + 3) <= 0
iii) Hence, find the range of (6x + 5) / (3x2 + 4x + 2)

b) Find the range of (x2 - 6x + 5) / (x2 + 2x + 1)

Well, for starters, you should write down the left hand side of the equation in i) and show that it is equal to zero.
• May 9th 2013, 06:06 PM
Re: Range and values of x
Attachment 28314

Excuse my handwriting, so I think I've solved i. How do I prove ii?
• May 9th 2013, 07:06 PM
Gusbob
Re: Range and values of x
Quote:

Attachment 28314

Excuse my handwriting, so I think I've solved i. How do I prove ii?

$3Dx^2 + (4D - 6)x + 2D - 5$ is a quadratic expression of form $ax^2+bx+c$ where $a=3D,b=(4D-6),c=(2D-5)$. How do you know when a quadratic equation $ax^2+bx+c=0$ has real solutions of $x$? HINT: look at the discriminant.
• May 9th 2013, 08:50 PM
Re: Range and values of x
Quote:

Originally Posted by Gusbob
$3Dx^2 + (4D - 6)x + 2D - 5$ is a quadratic expression of form $ax^2+bx+c$ where $a=3D,b=(4D-6),c=(2D-5)$. How do you know when a quadratic equation $ax^2+bx+c=0$ has real solutions of $x$? HINT: look at the discriminant.

Thanks Gusbob, however I'm stuck on this part. And how do I relate it back to "if and only if (D - 3)(2D + 3) <= 0"?
Attachment 28315
• May 9th 2013, 09:23 PM
Gusbob
Re: Range and values of x
Quote:

Under the square root sign, you made a mistake: should be $(4D-6)^2-...$ instead of $(4D+6)^2-...$
Notice that you get a quadratic term (which can be factorised) inside that square root. If you did your calculations correctly, it should read $-(D-3)(2D+3)$ inside the square root. You can get a real root if an only if this is a non-negative quantity, because you can't take (real) square roots of negative numbers.