The roots of x^2 + px + q =0 are doublet he roots of x^2 - (b+c)x + bc =0. Write p and q in terms of b and c.
Can someone show me how to do this please?
For $\displaystyle x^2 - (b+c)x + bc =0 $
Sum of roots is: $\displaystyle -\left(\frac{-(b+c)}{1}\right) = b+c $
Product of roots is: $\displaystyle \frac{bc}{1} = bc$
Clearly b and c are the roots of this equation (substitute into the equation and see). Therefore the roots of $\displaystyle x^2 + px + q =0 $ are 2b and 2c.
Sum of roots for this equation is $\displaystyle -\frac{p}{1} = -p $ therefore
$\displaystyle -p = 2b + 2c \Rightarrow p = -2(b+c) $
Product of roots is: $\displaystyle \frac{q}{1} = q $ therefore
$\displaystyle q = 2b \times 2c = 4bc $