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We thank Patrick from Woodbridge School for these solutions to parts (1) and (3).

**Update December 2013 Thanks to Tom from The Skinners School for sending in his solutions to parts (2) and (4). Scroll down to see it.

1) Let us take $y = 2x - 5 + \frac{3}{x}$, then we have $\frac{2x}{xy} + \frac {13x}{xy + 6x} = 6$, so $\frac {2}{y} + \frac {13}{y+6} = 6.$

Expanding the fractions, we have $2(y+6) + 13y = 6y(y+6) = 15y + 12.$ Dividing both sides by 3, we have $2y(y+6) = 5y + 4.$ Expanding the bracket, we have $2y^2 + 12y - 5y - 4 = 0 = 2y^2 + 7y - 4$. We can factorise this to give $(2y - 1)(y + 4) = 0$, so $y = \frac {1}{2}$ or $-4$.

We know $y = 2x - 5 + \frac{3}{x}$, so taking $y = \frac{1}{2}$ we have $\frac{1}{2} = 2x - 5 + \frac{3}{x}$ so we have $1 = 4x - 10 + \frac {6}{x}$ giving $4x^2 - 10x + 6 = x$ and then $4x^2 - 11x + 6 = 0.$ Factoring: $(4x - 3)(x - 2) = 0$, so $x = 2$ or $\frac{3}{4}$.

Now, take $y = -4$ then $-4 = 2x - 5 + \frac{3}{x}$ which gives $2x - 1 + \frac{3}{x} = 0 $ and $2x^2 - x + 3 = 0$ which has no real solutions.

Therefore, the solutions are $x = 2$ or $x = \frac{3}{4}$.

Alternatively, you could take advantage of the symmetry by substituting $y = 2x +\frac{3}{x}$ which simplifies the algebra a little. Patrick also sent a solution to part (3). Again substitutions turn the quartic equation into a quadratic which can be solved.

(3) $(x-4)(x-5)(x-6)(x-7) = 1680.$

Let us rewrite the equation with $a = x-7$, then we have $(a+3)(a+2)(a+1)a = 1680.$

Note that $a(a+3) = a^2 + 3a$, and $(a+2)(a+1) = a^2 + 3a + 2$, so $(a+2)(a+1)(a+3)a = (a^2 + 3a)(a^2 + 3a + 2).$ Let us then take $y = a^2 + 3a$, then $y(y+2) = 1680$ and $y^2 + 2y - 1680 = 0.$

Factorising, we find $(y-40)(y+42) = 0$, so $y = -42$ or $+40$. Since $y = a^2 + 3a$, then we have either: $40 = a^2 + 3a$, so $(a+8)(a-5) = 0$ and $a = -8$ or $5$.

Alternatively $-42 = a^2 + 3a$, which has no real roots. Therefore, we have $a = -8$ or $5$. Then, since $a = x - 7$, $x = a + 7$, so $x = -1$ or $12$.

We will test this by substitution: $(-1-4)(-1-5)(-1-6)(-1-7)$, so for each bracket (-a-b) we can take out the negative to give -(a+b). There are four such brackets, so there are four -1 factors removed, which multiply to give 1. Therefore, $(1+4)(1+5)(1+6)(1+7) = 5\times 6\times 7\times 8 = 1680.$ Similarly, $x=12$ gives us $8\times 7\times 6\times 5 = 1680.$

Well done Patrick. Again alternative substitutions include $t= x - \frac{11}{2}$ and a little more work will give the two complex solutions to this quartic equation, namely $\frac{11\pm i\sqrt159}{2}$.

Tom's solution to parts (2) and (4):

(2) $f(x) = x^4 - 8x^3 + 17x^2 - 8x + 1 = 0$

use substitution $z = x^2 - 8x + 1$

so $f(x) = zx^2 + 15x^2 +z = 0$

use same substitution again for the $x^2$ term that is multiplied with $z$

$f(x) = z(z^2 + 8x - 1) + 15x^2 + z = 0$

$= z^2 + 8xz - z + 15x^2 + z$

$= z^2 + 8xz + 15x^2$

$= (z+3x)(z+5x)$

so $x^2 - 5x + 1 = 0$

and $x^2 - 3x +1 =0$

you can solve these quadratics using the quadratic equation and this gives you the four solutions to f(x) = 0.

(4) $f(x) = (8x+7)^2(4x+3)(x+1) = \frac{9}{2}$

let $y = x+1$

$(8y-1)^2(4y-1)(y) = \frac{9}{2}$

$(64y^2-16y+1)(4y^2-y) =\frac{9}{2}$

let $z = 4y^2 - y$

$(16z+1)(z) = \frac{9}{2}$

$16z^2 + z =\frac{9}{2}$

$32z^2 + 2z - 9 = 0$

$(2z-1)(16z+9) = 0$

Second bracket gives no real values of $y$ so for real solutions, $2z - 1 =0$

$z = \frac{1}{2}$

$4y^2 - y = \frac{1}{2}$

$8y^2 -2y -1=0$

$(4y+1)(2y-1)=0$

$y= -\frac{1}{4} \text{and} \frac{1}{2}$

$x = -\frac{5}{4} \text{and} -\frac{1}{2}$