Bertie

I got the same result as you Reinout, and given part (b) of the question it was to be expected that the cubic had just one real root. (Differentiation of the quartic in (b) gets you 4 times the cubic).

750 + ('X% of 750 free') = 900.

X% of 750 = 150.

X = 20.

Almost Melody. I think that we still have to include x = 1/2 which makes the denominator zero.

$$\tan^{-1}(\frac{1}{\sqrt{x}})-\tan^{-1}(\sqrt{x})=105,$$

so, taking the tangent of both sides, and, using the identity

$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B},$$

together with

$$\tan(\tan^{-1}(\frac{1}{\sqrt{x}}))=\frac{1}{\sqrt{x}}\quad \text{ and }\quad \tan(\tan^{-1}(\sqrt{x}}))=\sqrt{x},$$

we have

$$\frac{1}{\sqrt{x}}-\sqrt{x}=2\tan 105.$$

Multiplying throughout by √x and rearranging,

$$x+2\sqrt{x}\tan 105-1=0,$$

which is a quadratic in √x.

Solving that, and taking the positive root gets √x≈7.595754.

To show that this satifies the original equation, remember that arctan is multivalued.

$$\tan^{-1}(\frac{1}{\sqrt{x}})-\tan^{-1}(\sqrt{x})=(7.5+k_{1}180)-(82.5+k_{2}180),$$

where k₁ and k₂ are integers.

 k₁ = 1 and k₂ = 0 produces the result 105, but there are an infinite number of other possibles.  

arctan(1/√x) and arctan(√x) are angles.

If you take the tangent of the equation and use the usual identity

tan(A - B) = (tan A - tan B)/(1 + tan A.tan B)

you finish up with a quadratic in √x.

I don't have time to type it in at the moment, I'll do so later if it hasn't been done in the meantime.

Also, is that 105 degrees or radians ? My guess is degrees.

Sorry Rom, but your answer would imply that the function is defined for, say, x = -5, which it isn't.

Equations like this need a numerical (iterative) method for their solution.

First though is to sketch a graph or graphs in order, if possible, to determine the number and approximate locations of the roots. For this particular example, best is to rewrite the equation as x² - 4x = 2^x and to sketch the graphs of y = x² - 4x, which is a concave up parabola crossing the x-axis at x=0 and x=4, and y = 2^x which looks rather like the y = e^x curve. At their point(s) of intersection x² - 4x = 2^x in which case their x co-ordinates will be the solutions to the original equation. There is just one intersection to the left of the origin, located (after trying two or three values for x), round about x = -0.2, and no intersections to the right, (the 2^x curve being already at a height of 16 as the parabola becomes positive at x=4).

The most popular numerical method for solving non-linear equations is probably the Newton-Raphson method.

For this equation the formula is

$$x_{n+1}=x_{n}-\frac{x^{2}_{n}-4x_{n}-2^{x_{n}}}{2x_{n}-4-2^{x_{n}}\ln(2)}$$

which, starting with x₀ = -0.2 converges to 0.206100 after just two iterations.

It's also usually possible to obtain a convergent iterative formula from the original equation. Rewriting the original equation as x = (x² - 2^x)/4 and using this as an iterative formula, (substitute x=-0.2 on the rhs to get x=-0.2076, substitute this into the rhs etc.), produces the sequence -0.2, -0.2076, -0.2057, -0.2062, -0.20608, etc. Not as fast as Newton-Raphson but it will get there eventually.

If you're multiplying, dividing or raising complex numbers to a power, it's best to work with polar form.

If z₁=r₁∠θ₁ and z₂=r₂∠θ₂, then the rules are, for multiplication, z₁z₂=r₁r₂∠(θ₁+θ₂), for division z₁/z₂=r₁/r₂∠(θ₁-θ₂) and raising to a power, zⁿ=rⁿ∠nθ.

So, (1+i)¹⁶=(√2∠(π/4))¹⁶=(√2)¹⁶∠(4π),

(1-i)¹⁵=(√2∠(-π/4))¹⁵=(√2)¹⁵∠(-15π/4), 

and ∴ (1+i)¹⁶/(1-i)¹⁵=((√2)¹⁶∠(4π))/((√2)¹⁵∠(-15π/4))=√2∠(4π+15π/4)=√2∠(-π/4),

and switching back to algebraic form, the result is 1-i.

Are you allowed to quote the Maclaurin series for cos(x) ? If yes then a simple substitution will do.

$$\cos x = 1 - \frac{x^{2}}{2!}+\frac{x^{4}}{4!} - \dots$$ ,

so, replacing x by x^2,

$$\cos x^{2}= 1 - \frac{x^{4}}{2!}+\frac{x^{8}}{4!}- \dots$$ .

I truly hate the result from the site calculator.

The result, x=4, is correct, but the presentation, saying that both sides of the original equation  are equal to x and equal to 4 is most certainly not correct.