Question

Derivation of quadratic formula by method of completing squares and also prove that

1. Sum of roots = -b/a
2. Product of roots = c/a
Hint : Standard form of quadratic equation is ax² + bx + c = 0​

Answer 1

We will solve this in steps to understand better! :)

Solution :

We have
`\begin{gathered}\sf a {x}^(2) + bx + c = 0 \end{gathered}` here a ≠ 0

Step 1 : Subtract 'c' from both sides of this equation

`\begin{gathered} \implies \sf a {x}^(2) + bx + c - c = - c \end{gathered}`

`\begin{gathered} \implies \sf a {x}^(2) + bx= - c \end{gathered}`

Step 2 : Dividing both side by coefficient of x² i.e 'a' [why? because we want the coefficient of x² as 1]

`\begin{gathered} \implies \sf \frac{ a {x}^(2)}{a} + (bx)/(a) = - (c)/(a) \end{gathered}`

`\begin{gathered} \implies \sf {x}^(2) + (bx)/(a) = - (c)/(a) \end{gathered}`

Step 3 : Adding
`\begin{gathered} { \bigg((1)/(2) * {\sf{coefficient}} \: x \bigg)}^(2) \end{gathered}` i.e,

`\begin{gathered}{ \bigg((1)/(2) * (b)/(a) \bigg)}^(2) \end{gathered}`

`\begin{gathered}{ \bigg( (b)/(2a) \bigg)}^(2) \end{gathered}` to both sides

`\begin{gathered} \implies \sf {x}^(2) + (bx)/(a) +{ \bigg( (b)/(2a) \bigg)}^(2) = - (c)/(a) +{ \bigg( (b)/(2a) \bigg)}^(2) \end{gathered}`

Step 4 : From the left side of equation an identity is formed i.e (a + b)² which is equal to a² + 2ab + b²

`\begin{gathered} \implies \sf { \bigg( x + (b)/(2a) \bigg)}^(2) = - (c)/(a) +{ \bigg( (b)/(2a) \bigg)}^(2) \end{gathered}`

Note : If we expand
`{ ( x + (b)/(2a) )}^(2)` it will again form
`{x}^(2) + (bx)/(a) +{ \bigg( (b)/(2a) \bigg)}^(2)`

Step 5 : Solving!

`\begin{gathered} \implies \sf { \bigg( x + (b)/(2a) \bigg)}^(2) = - (c)/(a) + \frac{ {b}^(2) }{4 {a}^(2) } \end{gathered}`

`\begin{gathered} \implies \sf { \bigg( x + (b)/(2a) \bigg)}^(2) = \frac{ {b}^(2) }{4 {a}^(2) } - (c)/(a) \end{gathered}`

Here, [On right side of equation for LCM]

4a² = 4×a×a

a = a

Hence LCM = 4a²

`\begin{gathered} \implies \sf { \bigg( x + (b)/(2a) \bigg)}^(2) = \frac{ {b }^(2) - 4ac }{ 4{a}^(2) } \end{gathered}`

`\begin{gathered} \implies \sf { \bigg( x + (b)/(2a) \bigg)}^(2) = \frac{ {b }^(2) - 4ac }{ {(2a)}^(2) } \end{gathered}`

`\begin{gathered} \implies \sf x + (b)/(2a) = \pm \sqrt{\frac{ {b }^(2) - 4ac }{ {(2a)}^(2) }} \end{gathered}`

`\begin{gathered} \implies \sf x = - (b)/(2a) \pm \sqrt{\frac{ {b }^(2) - 4ac }{ {(2a)}^(2) }} \end{gathered}`

`\begin{gathered} \implies \sf x = - (b)/(2a) \pm {\frac{\sqrt{ {b }^(2) - 4ac }}{ 2a }} \end{gathered}`

`\begin{gathered} \implies \boxed{ \sf x = {\frac{ - b \pm \sqrt{ {b }^(2) - 4ac }}{ 2a }}} \end{gathered}`

Our quadratic formula is formed!

Therefore,

The roots of general quadratic equation are

`\begin{gathered} \sf \alpha = {\frac{ - b + \sqrt{ {b }^(2) - 4ac }}{ 2a }} \end{gathered}`

`\begin{gathered} \sf \beta = {\frac{ - b - \sqrt{ {b }^(2) - 4ac }}{ 2a }} \end{gathered}`

Sum of roots
`\alpha + \beta`

`\begin{gathered} \implies \sf {\frac{ - b + \sqrt{ {b }^(2) - 4ac }}{ 2a }} + {\frac{ - b - \sqrt{ {b }^(2) - 4ac }}{ 2a }} \end{gathered}`

`\begin{gathered} \implies \sf {\frac{- b \: \cancel{+ \sqrt{ {b }^(2) - 4ac }} - b \: \cancel{- \sqrt{ {b }^(2) - 4ac }}}{ 2a }} \end{gathered}`

`\begin{gathered} \implies \sf {(- b - b )/( 2a )} \end{gathered}`

`\begin{gathered} \implies \sf {( - 2b )/( 2a )} \end{gathered}`

`\begin{gathered} \implies \boxed{ \sf {( - b )/( a )} }\end{gathered}`

So, the sum of roots = -b/a

Now,

Product of roots =
`\alpha \beta`

`\begin{gathered} \implies \sf \bigg( {\frac{ - b + \sqrt{ {b }^(2) - 4ac }}{ 2a }} \bigg) \bigg({\frac{ - b - \sqrt{ {b }^(2) - 4ac }}{ 2a }} \bigg) \end{gathered}`

`\begin{gathered} \implies \sf \bigg( \frac{ \sf{\overbrace{{ (- b)}}^{{a}}\: + \: \overbrace{{\sqrt{ {b }^(2) - 4ac}}}^{{b}} }}{ 2a } \bigg) \bigg({\frac{ { \overbrace{{ (- b)}}^{{a}}\: - \: \overbrace{{\sqrt{ {b }^(2) - 4ac}}}^{{b}}}}{ 2a }} \bigg) \end{gathered}`

Using (a + b) (a - b) = a² - b²

`\begin{gathered} \implies \sf {\frac{ { (- b)}^(2) - {( \sqrt{ {b }^(2) - 4ac })}^(2) }{ 4 {a}^(2) }} \end{gathered}`

`\begin{gathered} \implies \sf {\frac{ { b}^(2) - ( {b}^(2) - 4ac) }{ 4 {a}^(2) }} \end{gathered}`

`\begin{gathered} \implies \sf {\frac{ { b}^(2) - {b}^(2) + 4ac }{ 4 {a}^(2) }} \end{gathered}`

`\begin{gathered} \implies \sf {\frac{ 4ac }{ 4 {a}^(2) }} \end{gathered}`

`\begin{gathered} \implies \boxed{ \sf {( c )/( a)}} \end{gathered}`

So, Product of roots = c/a

We are done with our solution! :D

Note :

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