Solution of this system of equations: $2a+2b=14$ and $a b = 12$

Question

I need help with solving the following system of equations:

$$\begin{cases}2a+2b=14\\a b = 12\end{cases}$$

Thanks in advance.

Answer 1

with $$b=\frac{12}{a}$$ we get (after dividing by $2$): $$a+\frac{12}{a}=7$$ multplying by $a$: $$a^2-7a+12=0$$

Answer 2

This is the classical problem of finding two. It is a by-product of the theory of quadratic equations numbers, given their sum $s$ and their product $p$. If there exist two such numbers, they're the roots of the quadratic equation: $$x^2-sx+p=0.$$

Answer 3

1)$a+b= 7;$

2)$ab=12;$

$(a+b)^2 =a^2 +2ab +b^2=49;$

$(a-b)^2 = a^2-2ab+b^2 =$

$ (a+b)^2 -4ab = 49-48=1$.

3)$a-b = ^+_-1.$

Now add and subtract eqs. 1) and 3).

Note: $2a +2b= 14$ and $ab = 12 $ can be interpreted as the circumference and area of a rectangle with sides a,b.