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# Maths Equation Help

6y, 1m agoPosted 6 years, 1 month ago
I'm stuck on a piece of work, I've done everything apart from the very last bit.

I need to prove something...

I have to show the equation...

( (x-1)^2 ) / ( (sqrt(3)) * (sqrt(x^4 + 4x^2 +1)) )

Is always between 0 andf 1 when x isn't 1.

In other words the denominator is always greater than the numerator.

It's quite clear and I can show it graphically, but I need a proof for it which I'm stuck with.

When I say proof, it's not going to be anything complex at all.

James
6y, 1m agoPosted 6 years, 1 month ago
Options

(10)
#1
:-s ???
#2
It's for my Mechanics, got to prove two lines always intersect at an acute angle apart from when x=1

The equation is equal to cos(x) which needs to equal between 0 and 1 for the above statement to be true...

But not sure that will make it any clearer haha!
#3
Far too intelligent for people on here mate :-p
2 Likes #4
When 'x' is 1, the numerator ((x-1)^2) is 0, hence the result of the equation is 0.

However, as 'x' increases above 1, the denominator ( (sqrt(3)) * (sqrt(x^4 + 4x^2 +1)) ) tends towards approximately double of the numerator (from x = 16 onwards the denominator is just slightly more than double of the numerator).

Hence, the result of the equation falls in the range 0.5 to 0.6 (to 1 decimal place).

As you are obviously aware, the Cosine of an angle is the ratio of the length of the Adjacent side to the length of the Hypotenuse.

Can you "prove" a Cosine of between 0.5 & 0.6 is an acute angle (one less than 90 degrees)?

BFN,

fp.
#5
fanpages1 person likes this
When 'x' is 1, the numerator ((x-1)^2) is 0, hence the result of the equation is 0.

However, as 'x' increases above 1, the denominator ( (sqrt(3)) * (sqrt(x^4 + 4x^2 +1)) ) tends towards approximately double of the numerator (from x = 16 onwards the denominator is just slightly more than double of the numerator).

Hence, the result of the equation falls in the range 0.5 to 0.6 (to 1 decimal place).

As you are obviously aware, the Cosine of an angle is the ratio of the length of the Adjacent side to the length of the Hypotenuse.

Can you "prove" a Cosine of between 0.5 & 0.6 is an acute angle (one less than 90 degrees)?
.

Thank you very much for the reply!

I don't need to prove the cosine bit at all, just the first bit. Is there a way to prove algebraically what I've highlighted in bold as opposed to stating it? That's the bit I'm stuck on, I've got to show it for every positive and negative value of x that the numerator can never be greater than the denominator.

Cheers
James
#6
Hello
I started doing somethings with your equation but due to lack of time and no further clue I am unable to go further at this time. May be this will help you. try this if it helps.

( (x-1)^2 ) / ( (sqrt(3)) * (sqrt(x^4 + 4x^2 +1)) )

Let us say that this is = n

( (x-1)^2 ) / ( (sqrt(3)) * (sqrt(x^4 + 4x^2 +1)) ) = n

Do a power of 2 on both sides

( ( (x-1)^2 ) / ( (sqrt(3)) * (sqrt(x^4 + 4x^2 +1)) ) )^ 2 = n ^ 2

( (x-1)^2 ) ^ 2 / ( (sqrt(3)) * (sqrt(x^4 + 4x^2 +1)) ^ 2 ) = n^2

( (x-1)^4 ) / ( (sqrt(3)) ^ 2 * (sqrt(x^4 + 4x^2 +1)) ^ 2 ) = n ^ 2

( (x-1)^4 ) / ( 3 * (x^4 + 4x^2 +1) ) = n ^ 2

( (x-1)^4 ) / ( 3x^4 + 12x^2 +3) ) = n ^ 2
#7
OMG this has made me feel so thick, think I will stick with helping my 11 year old do fractions.
#8
Goofeys Girl
OMG this has made me feel so thick, think I will stick with helping my 11 year old do fractions.

me too,i wish i never saw this grrrr
#9
I tried doing it by putting it equal to something to no avail.

I've handed in what I did so will just wait and see now. Thank a lot for the help though.

James
banned#10
this is a wind up isn´t it?