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Question Number 174950 by Mastermind last updated on 14/Aug/22

Have you seen this method of solving  quadratic problem?  x^2 −x−12=0  y′=±(√(b^2 −4ac))

$$\mathrm{Have}\:\mathrm{you}\:\mathrm{seen}\:\mathrm{this}\:\mathrm{method}\:\mathrm{of}\:\mathrm{solving} \\ $$$$\mathrm{quadratic}\:\mathrm{problem}? \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{y}'=\pm\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}} \\ $$

Answered by peter frank last updated on 14/Aug/22

(d/dx)(x^2 −x−12)=±(√(b^2 −4ac))   2x−1=±(√(1^2 −4×−12))  2x−1=±(√(1+48))  2x−1=±(√(49))  2x−1=±7  2x=1±7  x=((1±7)/2)

$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{12}\right)=\pm\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}\: \\ $$$$\mathrm{2x}−\mathrm{1}=\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}×−\mathrm{12}} \\ $$$$\mathrm{2x}−\mathrm{1}=\pm\sqrt{\mathrm{1}+\mathrm{48}} \\ $$$$\mathrm{2x}−\mathrm{1}=\pm\sqrt{\mathrm{49}} \\ $$$$\mathrm{2x}−\mathrm{1}=\pm\mathrm{7} \\ $$$$\mathrm{2x}=\mathrm{1}\pm\mathrm{7} \\ $$$$\mathrm{x}=\frac{\mathrm{1}\pm\mathrm{7}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 14/Aug/22

Great sir.

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

Commented by Tawa11 last updated on 14/Aug/22

x = ((1 ± 7)/2)  x = 4 or − 3

$$\mathrm{x}\:=\:\frac{\mathrm{1}\:\pm\:\mathrm{7}}{\mathrm{2}} \\ $$$$\mathrm{x}\:=\:\mathrm{4}\:\mathrm{or}\:−\:\mathrm{3} \\ $$

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