Alan

You have the calculator set in degree mode if you get the 1.047... result.

Set in radian mode it gives the following:

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After t hours A will have walked 5t km north.

After t hours B will have cycled 16(t-1) km east.

The distance between them is given by Pythagoras:   sqrt( 5²t² + 16²(t-1)² ) → sqrt(281t² - 512t + 256) km

So we have:  sqrt(281t² - 512t + 256) = 52

Square both sides 

281t² - 512t + 256 = 2704

281t² - 512t - 2448 = 0

(281t + 612)(t - 4) = 0

The only positive solution is t = 4 hours.

Hence they are 52 km apart at 7 + 4 → 11 am.

Hmm.

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Guest #2 has the right idea, but has made a mistake in his cross-multiply stage.

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This is just (999999 - 999991)^2 = 8^2 → 64

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Note that (a - b)^2 = a^2 + b^2 - 2*a*b

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|2y-3|=29 

(1).  2 y > 3

       2y - 3 = 29

        2y = 32

        y = 16

(2).  2y < 3

       3 - 2y = 29 

       -2y = 26

         y = -13

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The result is zero:

\(\tan(x+45°)\rightarrow\frac{1+\tan x}{1-\tan x}\\\cot(x-45°)\rightarrow \frac{1+\tan x}{-1+\tan x}=-\tan(x+45°)\)

More generally:

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With a root of -1/2 it is likely that 2x^2 + 15x + p = 0 can be written as:

(2x + 1)*(x - a) = 0   This has roots of -1/2 and a

Expanding this we get:   2x^2 +(1 - 2a)*x - a = 0

For this to match we must have:  1 - 2a = 15  or  a = -7

We must also have  p = -a  so p = 7

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