Value of $a$ & $b$ individually

Question

If $a+b =5, ab = 6 $ what will be the proper way to find the individual value of $a$ and $b$? PS. I am not asking for hit & trial method.

Answer 1

Use a substitution. If $a+b=5$ and $ab=6$, then you know that $b=5-a$, and so $a(5-a)=6$. This is simply a quadratic, and can be solved using factoring or the quadratic formula: $$a(5-a)=6\iff a^2-5a+6=0\iff a=2\space\text{or}\space 3$$ Thus, we can see that the only two solution pairs $(a,b)$ are $(2,3)$ and $(3,2)$.

Answer 2

Remark that $a,b$ are solutions of the polynomial $(x-a)(x-b)=0$

When you develop it, this gives: $x^2-(a+b)x+ab=0$

So we are calling $\begin{cases} s=a+b &\text{the sum of a and b}\\ p=ab &\text{the product of a and b}\end{cases}$

Then $a,b$ are solutions of $$\boxed{x^2-sx+p=0}$$

Applying it to $s=5$ and $p=6$, we solve $x^2-5x+6=0$

$\Delta=25-4\times 6=1$ and $x=\dfrac{5\pm \sqrt{1}}2=3\text{ or }2$ and find $a,b$.

The formula with $s$ and $p$ is easy to remember and is very useful even in advanced problems.

Answer 3

1) Solve for one variable in terms of the other. You have four options.

i $a = 5-b;$ ii $b = 5-a;$ iii $a = \frac 6b;$ iv) $b = \frac 6a$.

2) Plug that into the other equation:

i $(5-b)b = 6$; ii $a(5-a)= 6;$ iii $\frac 6b + b = 5$ iv $a + \frac 6a = 5$.

3) Solve:

i $b^2 - 5b + 6=0$ so $b = 2,3$.

ii) $a^2 - 5a + 6 = 0$ so $a = 2,3$.

iii) $b^2 - 5b + 6 = 0$ so $b = 2,3$

iv) $a^2 -5a + 6 = 0$ so $a = 2,3$.

Plug into first expression

i $a = 5- b$ if $b =2$ then $a =3$ if $a = 3$ then $a =2$. So solutions are $(2,3), (3,2)$.

ii) $b = 5-1$ if $a= 2$ then $b = 3$ if $a = 2$ then $b= 3$.So solutions are $(2,3), (3,2)$.

iii) $a = \frac 6b$ if $b =2$ then $a =3$ if $a = 3$ then $a =2$. So solutions are $(2,3), (3,2)$.

iv) $b = \frac 6a$ if $a= 2$ then $b = 3$ if $a = 2$ then $b= 3$.So solutions are $(2,3), (3,2)$.

Whether you want to solve via i, ii, iii, or iv the choice is up to you.