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Target Six

Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.

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This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?

Complex Partial Fractions

Age 16 to 18
Challenge Level

Thank you Joseph O'keefe from Colyton Grammar School for this solution. Andrei Lazanu from Rumania and Laura Hannick, Townley Grammar School for Girls, also sent in good solutions.

Part (1)

$${10x^2-2x+4\over x^3 + x} = {A\over x} + {Bx+C\over x^2+1}$$ In order to work out the constants $A,B,C$, I put the RHS of the identity over a common denominator: $${10x^2-2x+4\over x^3 + x} = {A(x^2+1)+ x(Bx+C)\over x(x^2+1)}$$ Since the denominators are equal, it follows that the numerators of the fraction must also be equal: $$\eqalign{ 10x^2-2x+4 &= A(x^2+1)+ x(Bx+C)\cr &= Ax^2+Bx^2+Cx+A \cr &= (A+B)x^2+Cx+A}$$ By comparing coefficients, it follows that $A+B=10$, $C=-2$ and $A=4$ so $B=6$.

Part (2)

$${A\over x} + {Bx+C\over x^2+1}={A\over x} + {D\over (x-i)} + {E\over (x+i)}.$$ By taking $A/x$ away and replacing $B$ and $C$ by their respective values, and putting $D$ and $E$ over a common denominator: $${6x-2\over (x^2+1)}= {D(x+i)+E(x-i)\over (x-i)(x+i)}.$$ Again, since the denominators are equal, it follows that the numerators are equal so $6x-2=D(x+i)+E(x-i)$.

By comparing coefficients we have $D+E=6$ and $(D-E)i=-2$. Then $6i-2=D(2i)$ so $D=(6i-2)/2i$. Multiply both the numerator and the denominator by $i$ to get a real denominator: $$D={6i^2-2i\over -2} = {-6-2i\over -2} = 3+i$$ Then $E(-2i)= -6i-2$, so $E=3-i$. So $A=4$, $B=6$, $C=-2$, $D=3+i$ and $E=3-i$ $${10x^2-2x+4\over x^3 + x}= {4\over x} + {6x-2\over x^2+1}={4\over x} + {3+i\over (x-i)} + {3-i\over (x+i)}$$