\therefore [\tex] when [tex] b^2 < 4ac [/tex] f(x) > 0 for all real values of x [br][br][br][br]Please I want to know if my proof correctly answers the question please tell where i went wrong [br]thank you
\displaystlye f(x) = a \underbrace{\left ( x+\frac{b}{2a}\right )^2}_{\geq 0} + \displaystyle \frac{4ac-b^2}{4a}
\displaystlye f(x) = a \underbrace{\left ( x+\frac{b}{2a}\right )^2}_{\geq 0} + \displaystyle \frac{4ac-b^2}{4a}
\a\left ( x+\frac{b}{2a}\right )^2 \geq 0
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong71
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrong71