Re: Range and values of x

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**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) / (3x^{2 }+ 4x + 2)

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

i) Show that 3Dx^{2} + (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) / (3x^{2} + 4x + 2)

b) Find the range of (x^{2 }- 6x + 5) / (x^{2} + 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.

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Re: Range and values of x

Attachment 28314

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

Re: Range and values of x

Quote:

Originally Posted by

**Caddym89** Attachment 28314
Excuse my handwriting, so I think I've solved i. How do I prove ii?

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

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Re: Range and values of x

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**Gusbob** $\displaystyle 3Dx^2 + (4D - 6)x + 2D - 5$ is a quadratic expression of form $\displaystyle ax^2+bx+c$ where $\displaystyle a=3D,b=(4D-6),c=(2D-5)$. How do you know when a quadratic equation $\displaystyle ax^2+bx+c=0$ has real solutions of $\displaystyle x$? 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

Re: Range and values of x

Quote:

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"?

Attachment 28315

Under the square root sign, you made a mistake: should be $\displaystyle (4D-6)^2-...$ instead of $\displaystyle (4D+6)^2-...$

Notice that you get a quadratic term (which can be factorised) inside that square root. If you did your calculations correctly, it should read $\displaystyle -(D-3)(2D+3)$ 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.