# Math Help - Range and values of x

1. ## 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)

2. ## Re: Range and values of x

Originally Posted by Caddym89
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.

3. ## Re: Range and values of x

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

4. ## Re: Range and values of x

Originally Posted by Caddym89

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.

5. ## Re: Range and values of x

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"?

6. ## Re: Range and values of x

Originally Posted by Caddym89
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"?
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.