You may also like

Roots and Coefficients

If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?

Three by One

There are many different methods to solve this geometrical problem - how many can you find?

Target Six

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

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 $E=3-i$ $${10x^2-2x+4\over x^3 + x}= {4\over x} + {6x-2\over x^2+1}={4\over x} + {3+i\over (x-i)} + {3-i\over (x+i)}$$