Question

obtain quadratic formula from general form of quadratic equation​

Answer 1

Answer:

No Final answer. It is a derivation .

Step-by-step explanation:

ax² +bx + c = 0 (general form of quad. equation)

Taking out a common.

a (x²+bx/a+c/a) = 0

Transposing a to RHS .

x²+(bx/a) + (c/a) = 0

Multiplying numerator and denominator of middle term i.e bx/a by 2

x²+(2bx/2a) + (c/a) = 0

Taking out 2 from brackets.

x² +2(bx/2a) +(c/a) = 0

Adding & subtracting (b/2a)² on LHS ,

x² +2(bx/2a) +(b/2a)² - (b/2a)² +(c/a) = 0

Now , first three terms are of the form a²+2ab+b². So, we can write is as (a+b)².

Now , first three terms are of the form a²+2ab+b². So, we can write is as (a+b)².Here a = x & b = b/2a ( Key Step)

${(x + \frac{b}{2a} )}^{2} - {( { \frac{b}{2a} } )}^{2} + \frac{c}{a} = 0$

Transposing -(b/2a)² + (c/a) to RHS.

We get :

${(x + \frac{b}{2a} )}^{2} = {( \frac{b}{2a} )}^{2} - \frac{c}{a}$

${(x + \frac{b}{2a} )}^{2} = \frac{ {b}^{2} }{4 {a}^{2} } - \frac{c}{a}$

Taking LCM on RHS.

We get :

${(x + \frac{b}{2a} )}^{2} = \frac{ {b}^{2} - 4ac}{4 {a}^{2} }$

Now , taking square root of both sides of the equation .

We get :

x + b/2a = ±√(b² - 4ac)/2a

x = (-b/2a) ± {√(b² - 4ac)/2a}

Adding terms on RHS , we get :

x = (-b ± √b²-4ac)/2a

Hence our quadratic formula is derived.

Answer 2

Your answer is :-

To obtained quadratic formula from general form , we use completing of square method .

See the attachment