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Question Number 9020 by tawakalitu last updated on 14/Nov/16

Commented by RasheedSoomro last updated on 20/Nov/16

By [((ax^2 +bx+c)/(2x+b))] do you mean bracket function?

$$\mathrm{By}\:\left[\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{2x}+\mathrm{b}}\right]\:\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{bracket}\:\mathrm{function}? \\ $$

Answered by Rasheed Soomro last updated on 14/Nov/16

x^([((ax^2 +bx+c)/(2x+b))]) =1  x^([((ax^2 +bx+c)/(2x+b))]) =x^0   [((ax^2 +bx+c)/(2x+b))]=0  0≤((ax^2 +bx+c)/(2x+b))<1  Continue

$$\mathrm{x}^{\left[\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{2x}+\mathrm{b}}\right]} =\mathrm{1} \\ $$$$\mathrm{x}^{\left[\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{2x}+\mathrm{b}}\right]} =\mathrm{x}^{\mathrm{0}} \\ $$$$\left[\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{2x}+\mathrm{b}}\right]=\mathrm{0} \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{2x}+\mathrm{b}}<\mathrm{1} \\ $$$$\mathrm{Continue} \\ $$

Commented by tawakalitu last updated on 14/Nov/16

thank sir.

$$\mathrm{thank}\:\mathrm{sir}. \\ $$

Answered by mrW last updated on 15/Nov/16

x^([(((ax^2 +bx+c))/((2x+b)))]) =1  x=1   or  ((ax^2 +bx+c)/(2x+b))=0  i.e.  ax^2 +bx+c=0  x=((−b±(√(b^2 −4ac)))/(2a))  condition: 2x+b≠0  ((−b±(√(b^2 −4ac)))/a)+b≠0  c≠((b^2 (2−a))/4)

$$\mathrm{x}^{\left[\frac{\left(\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}\right)}{\left(\mathrm{2x}+\mathrm{b}\right)}\right]} =\mathrm{1} \\ $$$$\mathrm{x}=\mathrm{1}\: \\ $$$$\mathrm{or} \\ $$$$\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{2x}+\mathrm{b}}=\mathrm{0} \\ $$$$\mathrm{i}.\mathrm{e}. \\ $$$$\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{−\mathrm{b}\pm\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}}{\mathrm{2a}} \\ $$$$\mathrm{condition}:\:\mathrm{2x}+\mathrm{b}\neq\mathrm{0} \\ $$$$\frac{−\mathrm{b}\pm\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}}{\mathrm{a}}+\mathrm{b}\neq\mathrm{0} \\ $$$$\mathrm{c}\neq\frac{\mathrm{b}^{\mathrm{2}} \left(\mathrm{2}−\mathrm{a}\right)}{\mathrm{4}} \\ $$

Commented by tawakalitu last updated on 15/Nov/16

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Commented by Rasheed Soomro last updated on 16/Nov/16

((ax^2 +bx+c)/(2x+b)) is not only equal to zero,but  also has any value between 0 and 1(excluded 1).

$$\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{2x}+\mathrm{b}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{only}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{zero},\mathrm{but} \\ $$$$\mathrm{also}\:\mathrm{has}\:\mathrm{any}\:\mathrm{value}\:\mathrm{between}\:\mathrm{0}\:\mathrm{and}\:\mathrm{1}\left(\mathrm{excluded}\:\mathrm{1}\right). \\ $$

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