## Discussion Forum

### Daily magic problem July 9th

Daily magic problem July 9th

Using r1+r2 will give different answer from r1*r2. Why?

Re: Daily magic problem July 9th

The sum of the two roots $r_1 + r_2$ must be equal to $-(-5)/1 = 5$, and the product of the roots $r_1r_2$ must be equal to $7 + i$.

Since $(2 + i) + (a + ib) = (2 + a) + (1 + b)i = 5 + 0i$, we must have $2 + a = 5$ and $1 + b = 0$. Similarly, since $(2 + i)(a + ib) = (2a - b)(a + 2b)i = 7 + i$, we must have $2a - b = 7$ and $a + 2b = 1$.

Both systems of equations lead to the same solution $(a,b)$.

Re: Daily magic problem July 9th

Thank you for your explanation! How about using z*(2+i)=7+i Then z=(7+i)/(2+i) This is different. Also, is it ok to use quadratic formula to find solutions? Is it the same?

Re: Daily magic problem July 9th

Note your solution $z = \dfrac{7 + i}{2 + i}$ is the same as $3 - i$: $$\dfrac{7 + i}{2 + i} = \dfrac{7 + i}{2 + i}\dfrac{2-i}{2-i} = \dfrac{15 - 5i}{5} = 3 - i.$$

Using the quadratic formula is always an option, though it can get messy and you wouldn't be taking advantage of the fact that you were given one of the two solutions already.

Re: Daily magic problem July 9th

Thank you!