# Thread: Is there a simpler solution?

1. ## Is there a simpler solution?

Hi

Relation
$\displaystyle{\frac{(500+x)}{700} = \frac{c}{40}}$

and
$\sqrt{c^2+40^2}=40+x$

$c=\sqrt{x²+80x}$

gives
$\displaystyle{\frac{(500+x)}{700}=\frac{\sqrt{x²+8 0x}}{40}}$

How do I solve above?

2. Originally Posted by pelle2004
Hi

Relation
$\displaystyle{\frac{(500+x)}{700} = \frac{c}{40}}$

and
$\sqrt{c^2+40^2}=40+x$

$c=\sqrt{x²+80x}$

gives
$\displaystyle{\frac{(500+x)}{700}=\frac{\sqrt{x²+8 0x}}{40}}$

How do I solve above?

When dealing with real numbers, this logic might be handy:

$\displaystyle $a = \sqrt b \Rightarrow \left\{ \begin{gathered} {a^2} = b \hfill \\ b \geqslant 0 \hfill \\ a \geqslant 0 \hfill \\ \end{gathered} \right.$$

3. Originally Posted by pelle2004
Hi

Relation
$\displaystyle{\frac{(500+x)}{700} = \frac{c}{40}}$

and
$\sqrt{c^2+40^2}=40+x$

$c=\sqrt{x²+80x}$

gives
$\displaystyle{\frac{(500+x)}{700}=\frac{\sqrt{x²+8 0x}}{40}}$

How do I solve above?

Square both sides to get a quadratic. Solve the quadratic and check the solutions by substituting back into the above to make sure you have not introduced spurious solutions by squaring.

CB

4. Not shure what I should do by 'quadratic'

$\displaystyle{\frac{500+x}{17,5}}=\sqrt{x²+80x}$

$\displaystyle{\frac{500²+x²+1000x}{17,5²}}=x²+80x$

$x²+80x-\displaystyle{\frac{500²}{17,5²}-\frac{x²}{17,5²}-\frac{1000}{17,5²}x}=0$

$(17,5²-1)x²+(17,5²*80-1000)x-500²=0$

$305,25x²+23500x-250000=0$

$x²+\displaystyle{\frac{23500}{305,25}}x=250000$

I guess....

$y(x)=\frac{250000*305,25}{23500}+Ce^{-\frac{23500}{305,25}x}$

But then I need some help to be able to solve x for y=0 ...

5. Originally Posted by pelle2004
Not shure what I should do by 'quadratic'

$\displaystyle{\frac{500+x}{17,5}}=\sqrt{x²+80x}$

$\displaystyle{\frac{500²+x²+1000x}{17,5²}}=x²+80x$

$x²+80x-\displaystyle{\frac{500²}{17,5²}-\frac{x²}{17,5²}-\frac{1000}{17,5²}x}=0$

$(17,5²-1)x²+(17,5²*80-1000)x-500²=0$

$305,25x²+23500x-250000=0$
The last quoted line is OK.

A quadratic equation

$a x^2+bx +c = 0$ has the solution $x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

you know: a = 305.25, b = 23500, c = -250000

so you can calculate the value of x. (Btw the negative solution isn't very plausible with your question)

6. Originally Posted by pelle2004
$\displaystyle{\frac{(500+x)}{700}=\frac{\sqrt{x²+8 0x}}{40}}$
Less confusion if next is rearranging this way:

35SQRT(x^2 + 80x) = 2x + 1000 ; then:

1225x^2 + 98000x = 4x^2 + 4000x + 1000000

1221x^2 + 94000x - 1000000 = 0

7. Originally Posted by earboth
The last quoted line is OK.

A quadratic equation

$a x^2+bx +c = 0$ has the solution $x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

you know: a = 305.25, b = 23500, c = -250000

so you can calculate the value of x. (Btw the negative solution isn't very plausible with your question)
x=9.4727287939

/Thanks!!