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Question Number 14719 by Abbas-Nahi last updated on 03/Jun/17

Commented by Abbas-Nahi last updated on 03/Jun/17

Is the solution possible mathmatically orNOT and whether it is logical or not

$${Is}\:{the}\:{solution}\:{possible}\:{mathmatically} \\ $$$${orNOT}\:{and}\:{whether}\:{it}\:{is}\:{logical}\:{or}\:{not} \\ $$$$ \\ $$

Commented by Abbas-Nahi last updated on 03/Jun/17

really want to help?

$${really}\:{want}\:{to}\:{help}? \\ $$

Commented by mrW1 last updated on 04/Jun/17

4−(9/2)≠(√((4−(9/2))^2 )) 4−(9/2)=−(√((4−(9/2))^2 ))=−(√((5−(9/2))^2 ))=−(5−(9/2))

$$\mathrm{4}−\frac{\mathrm{9}}{\mathrm{2}}\neq\sqrt{\left(\mathrm{4}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}−\frac{\mathrm{9}}{\mathrm{2}}=−\sqrt{\left(\mathrm{4}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} }=−\sqrt{\left(\mathrm{5}−\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} }=−\left(\mathrm{5}−\frac{\mathrm{9}}{\mathrm{2}}\right) \\ $$

Commented by mrW1 last updated on 04/Jun/17

generally a≠(√a^2 )=∣a∣ a=(√a^2 )=∣a∣ if a≥0 a=−(√a^2 )=−∣a∣ if a<0

$${generally} \\ $$$${a}\neq\sqrt{{a}^{\mathrm{2}} }=\mid{a}\mid \\ $$$${a}=\sqrt{{a}^{\mathrm{2}} }=\mid{a}\mid\:{if}\:{a}\geqslant\mathrm{0} \\ $$$${a}=−\sqrt{{a}^{\mathrm{2}} }=−\mid{a}\mid\:{if}\:{a}<\mathrm{0} \\ $$

Commented by RasheedSoomro last updated on 04/Jun/17

a=(√a^2 )=a if a≥0 a=−(√a^2 )=−a if a<0 No need to use ∣ ∣ Am I right?

$${a}=\sqrt{{a}^{\mathrm{2}} }={a}\:{if}\:{a}\geqslant\mathrm{0} \\ $$$${a}=−\sqrt{{a}^{\mathrm{2}} }=−{a}\:{if}\:{a}<\mathrm{0} \\ $$$$\mathrm{No}\:\mathrm{need}\:\mathrm{to}\:\mathrm{use}\:\:\:\mid\:\:\:\:\mid \\ $$$$\mathrm{Am}\:\mathrm{I}\:\mathrm{right}? \\ $$

Commented by Tinkutara last updated on 04/Jun/17

Yes, in my opinion.

$$\mathrm{Yes},\:\mathrm{in}\:\mathrm{my}\:\mathrm{opinion}. \\ $$

Commented by mrW1 last updated on 04/Jun/17

you are right sir. I just want to say (√a^2 ) is only the absolute value of a number, but not always the number itself.

$${you}\:{are}\:{right}\:{sir}.\:{I}\:{just}\:{want}\:{to}\:{say}\:\sqrt{{a}^{\mathrm{2}} } \\ $$$${is}\:{only}\:{the}\:{absolute}\:{value}\:{of}\:{a}\:{number}, \\ $$$${but}\:{not}\:{always}\:{the}\:{number}\:{itself}. \\ $$

Commented by Abbas-Nahi last updated on 04/Jun/17

thank you for your cooperation with my best regards

$${thank}\:{you}\:{for}\:{your}\:{cooperation}\:{with} \\ $$$${my}\:{best}\:{regards} \\ $$

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