Define sridhar acharya rule
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6
Sridhar acharya rule related to quadratic equation
let any quadratic equation ,
ax^2+bx+c=0 where a isn't equal to zero
=> ax^2+bx=-c
=> x^2+b/ax=-c/a
=>x^2+b/ax+(b/2a)^2=-c/a+(b/2a)^2
=>(x+b/2a)^2=(b^2-4ac)/4a^2
take root both side
=>(x+b/2a)=+_root(b^2-4ac)/2a
= x= {-b +_root (b^2-4ac)}/2a
let any quadratic equation ,
ax^2+bx+c=0 where a isn't equal to zero
=> ax^2+bx=-c
=> x^2+b/ax=-c/a
=>x^2+b/ax+(b/2a)^2=-c/a+(b/2a)^2
=>(x+b/2a)^2=(b^2-4ac)/4a^2
take root both side
=>(x+b/2a)=+_root(b^2-4ac)/2a
= x= {-b +_root (b^2-4ac)}/2a
mysticd:
Abhi , plz correct,(x+b/2a)^2 before line
x^2+2*x*b/2a+....for user understanding
sorry I am too late for edit
Answered by
1
let any quadratic equation ,
ax^2+bx+c=0 where a isn't equal to zero
=> ax^2+bx=-c
=> x^2+b/ax=-c/a
=>x^2+b/ax+(b/2a)^2=-c/a+(b/2a)^2
=>(x+b/2a)^2=(b^2-4ac)/4a^2
take root both side
=>(x+b/2a)=+_root(b^2-4ac)/2a
ax^2+bx+c=0 where a isn't equal to zero
=> ax^2+bx=-c
=> x^2+b/ax=-c/a
=>x^2+b/ax+(b/2a)^2=-c/a+(b/2a)^2
=>(x+b/2a)^2=(b^2-4ac)/4a^2
take root both side
=>(x+b/2a)=+_root(b^2-4ac)/2a
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