the difference between two numbers a and b is 1, and the product of the numbers is 9.

Question

Full question: the difference between two numbers a and b is 1, and the product of the numbers is 9. Show that there are two possible pairs of values which satisfy these condictions and evaluate the numbers

This is not to do with complex numbers so you cannot have negative roots if using quadratic equation (which i seem to keep getting.)

Answer 1

you will get $$x-y=1$$ and $$xy=9$$ if $$x\geq y$$ thus we have the quadratic equation $$(1+y)y=9$$ and $$y^2+y-9=0$$ can you solve this?

Answer 2

Ok let's do a few steps: $$ \begin{cases} a-b = 1 \\ ab = 9 \end{cases} $$ The first equations tells you $a = 1+b$. Putting it into the second you get $$(1+b)b = 9 \implies b^2 + b = 9 $$ Solving this classic quadratic equation gives : $b_{1,2} = -\frac{1}{2} \pm \frac{\sqrt{37}}{2}$

Then you succesively substitute $b_1$ and $b_2$ in the first equation and you find the corresponding $a_1$ and $a_2$ which will give you two pairs of solutions $(a_1,b_1), (a_2,b_2)$.