Question

If $x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$; make ‘c’ as the subject of the formula.

Answer 1

Hint: When a variable or a coefficient is from one side of equal to sign to another side the signs get interchanged and multiplication becomes division and vice versa.

Complete step by step answer:
In order to make c as subject we should bring c on one side and other terms on other side
Firstly, we will bring 2a on the LHS
$\therefore 2ax = - b + \sqrt {{b^2} - 4ac} $
Now we will transpose -b on RHS and the sign of – with b will get interchanged to + as per the hint
$\therefore 2ax + b = \sqrt {{b^2} - 4ac} $
Now we will take square on both sides to remove the square root sign of the RHS
$\therefore {(2ax + b)^2} = {b^2} - 4ac$
Now we will bring ${b^2}$on the LHS and it will become -${b^2}$ as per the hint
$\therefore {(2ax + b)^2} - {b^2} = - 4ac$
Now we will bring -4a on the LHS and it will become $\dfrac{{ - 1}}{{4a}}$ as per the hint
$\therefore - \dfrac{{{{(2ax + b)}^2} - {b^2}}}{{4a}} = c$
Therefore, $c = \dfrac{{{{(2ax + b)}^2} - {b^2}}}{{4a}}$

Note:
While solving the equation $2ax + b = \sqrt {{b^2} - 4ac} $ , we basically want to get rid of square root sign and we successfully get rid of it by squaring both side of the equation because $\sqrt {{b^2} - 4ac} $ means ${({b^2} - 4ac)^{\dfrac{1}{2}}}$ and when we square, it becomes ${\left( {{{({b^2} - 4ac)}^{\dfrac{1}{2}}}} \right)^2} = {({b^2} - 4ac)^{\dfrac{1}{2} \times 2}}$ and product of $\dfrac{1}{2}$ and 2 is 1. Therefore, $\sqrt {{b^2} - 4ac} $ becomes ${b^2} - 4ac$.