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How these results have been derived ? I always to fail to memorise them so I thought why not go with the logic. Because these are widely in Differential Calculus too though they are part of Quadratic Eqautions

How these results have been derived ?
I always to fail to memorise them so I thought why not go with the logic.
 
 
Because these are widely in Differential Calculus too though they are part of Quadratic Eqautions

Question Image
Grade:12

1 Answers

Ajay
209 Points
4 years ago
What study material are you following, you should refer to a classic text series like Higher Algebra by Hall and Knight to clear such concepts.
Even iit jee books like one from Arihant got algebra have these concepts explained nicely.
Coming back to your query here is the explanation
Let\quad f(x)\quad =\quad a{ x }^{ 2 }+bx+c\\ \quad \quad \quad \quad \quad \quad \quad =\quad a\left( { x }^{ 2 }+\frac { b }{ a } x+\frac { c }{ a } \right) \\ \quad \quad \quad \quad \quad \quad \quad =\quad a\left[ { \left( { x }+\frac { b }{ 2a } \right) }^{ 2 }\quad +\quad c-\frac { { b }^{ 2 } }{ { 4a }^{ 2 } } \right] \\ \quad \quad \quad f(x)\quad =\quad a\left[ { \left( { x }+\frac { b }{ 2a } \right) }^{ 2 }\quad -\frac { D }{ { 4a }^{ 2 } } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad where\quad D\quad ={ \quad b }^{ 2 }-4ac
 
It is clear at the expression inside aquare brackets will be postive if D is less than 0.
Also if a is  postive and D is negative f(x) is postive.
SImilarly if a is negative and D is negtaive, f(x) is negative.
 

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