algebra question (1 Viewer)

[Loner]

Here's an algebra problem that I hope someone can help me with. Imagine you have a right angled triangle (height = h) and base = x. The perimeter is 12 units and the area is 7 square units.

Thus,

7 = 1/2.x.h (area) and

12 = h + x + sqrt (x^2 + h^2) (perimeter)

These two equations can be combined and should reduce to the quadratic equation in terms of x. I tried for ages and wasn't able to do it. Apparently the quadratic equation gives rise to a complex solution.

Thanks in advance..

 

[Riviet]

Hmm... this would probably be better in the extension 2 forum.
Anyway, from the area of the triangle formula, make h the subject, ie h=14/x, and substitute it into the perimeter equation to obtain:

14/x + x + sqrt(x²+196/x²) = 12

196/x² + x² + x² + 196/x² = 144

196 + x⁴ + x⁴ + 196 = 144

2x⁴=-248

x⁴=-124

As you can see, this will produce complex roots.

The only systematic way I can think of now of solving that is the equating the re and im parts of the expansion of (a+ib)⁴.

(a+ib)⁴ = -124+0i

Expanding LHS, and equating im and re parts we obtain:

a⁴ + b⁴ - 6a²b² = -124 [1]

4a³b - 4ab³ = 0 [2]

From [2],

ab(a+b)(a-b) = 0

ab=0 or a+b=0 (a=-b) or a-b=0 (a=b)

That's all I've done. I hope that helps a bit.

 

[Trev]

14/x + x + sqrt(x²+196/x²) = 12

196/x² + x² + x² + 196/x² = 144

If you're squaring something you have to square all of it not in parts, so that step is wrong, yea?

The third line is also incorrect.
You multiply by x² only the LHS, the RHS should be 144x² not 144.

 

[Riviet]

Thanks for spotting that Trev.

Might have another go at it then.

 

[insert-username]

7 = (x.h)/2 (area) (1)

12 = h + x + sqrt (x^2 + h^2) (perimeter) (2)


Therefore 14 = x.h

x = 14/h

12 = h + 14/h + √(196/h² + h²)

√(196/h² + h²) = 12 - h - 14/h

196/h² + h² = (12 - h - 14/h)²

196/h² + h² = 144 - 12h - 168/h - 12h + h² + 14 - 168/h + 14 +168/h²

196/h² + h² = h² - 24h + 172 - 336/h + 168/h²

0 = -24h + 172 - 336/h - 28/h²

0 = -24h³ + 172h² - 336h - 28

0 = -6h³ + 43h² - 84h - 7

From a quick substitution (f (-1) = 40, f(0) = -7) this function has one root between -1 and 0. But h is a distance, so it must be a positive root. I've stuffed up somewhere.... or it's totally right and I'm looking for a complex root that I can't be bothered to find. Excellent.


I_F

 

[Riviet]

sqrt(x²+196/x²)=12-[(x²+14)/x]

Squaring and expanding LHS and RHS,

x²+196/x²=144-(24x²+336)/x+(x⁴+28x²+196)/x²

Removing denominators,

x⁴+196=144x²-(24x³+336x)+x⁴+28x²+196

24x³-172x²+336x=0

6x³-43x²+84x=0

x(6x²-43x+84)=0 [reject x=0 since it is the base length]

By the quadratic formula, x=43/12 + sqrt(167).i/12

Then substitute x into h=14/x to find h.

That should be right now.

 

[Trev]

Riviet:
I got the same answer as you. Gw.

 

[Riviet]

Nice work.

 

[Loner]

Thanks guys, I spotted my mistake. It was with the expansion part. I should have seen that (a+b+c)^2 is the same as ((a+b) + c)^2. Maybe I should go back to grade 8 again!