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Question Number 8600 by Chantria last updated on 17/Oct/16

Test 1=(√1)=(√((−1)(−1)))=(√(−1))∙(√(−1)) =i∙i=i^2 =−1 so 1=−1 Find the error.

$$\:\boldsymbol{{Test}}\: \\ $$$$\:\mathrm{1}=\sqrt{\mathrm{1}}=\sqrt{\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)}=\sqrt{−\mathrm{1}}\centerdot\sqrt{−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={i}\centerdot{i}={i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\:\:{so}\:\mathrm{1}=−\mathrm{1} \\ $$$$\:{Find}\:{the}\:{error}. \\ $$

Commented by prakash jain last updated on 17/Oct/16

(ab)^x =a^x ∙b^x only for a,b,x∈R and a,b,x are +ve. (−1)^(1/2) (−1)^(1/2) ≠(−1×−1)^(1/2) since −1<0

$$\left({ab}\right)^{{x}} ={a}^{{x}} \centerdot{b}^{{x}} \: \\ $$$$\mathrm{only}\:\mathrm{for}\:{a},{b},{x}\in\mathbb{R}\:\mathrm{and}\:{a},{b},{x}\:\mathrm{are}\:+\mathrm{ve}. \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} \left(−\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} \neq\left(−\mathrm{1}×−\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} \:\mathrm{since}\:−\mathrm{1}<\mathrm{0} \\ $$

Answered by malwaan last updated on 18/Oct/16

(√1)=±1 so 1=(√1) is one part from the solution and (√1)=−1 is the 2nd part 1=(√1)=−1 is wrong (√1)=±1 is right

$$\sqrt{\mathrm{1}}=\pm\mathrm{1}\: \\ $$$${so}\:\mathrm{1}=\sqrt{\mathrm{1}}\:{is}\:{one}\:{part}\:{from}\:{the}\:{solution} \\ $$$${and}\:\sqrt{\mathrm{1}}=−\mathrm{1}\:{is}\:{the}\:\mathrm{2}{nd}\:{part} \\ $$$$\mathrm{1}=\sqrt{\mathrm{1}}=−\mathrm{1}\:{is}\:{wrong} \\ $$$$\sqrt{\mathrm{1}}=\pm\mathrm{1}\:{is}\:{right} \\ $$

Commented by Rasheed Soomro last updated on 19/Oct/16

“ (√( )) ” is used only for +ve root. Hence (√1) = 1 not (√1)=±1 We can write ±(√1)=±1 but we can′t write (√1)=±1

$$``\:\sqrt{\:\:}\:''\:\mathrm{is}\:\mathrm{used}\:\mathrm{only}\:\mathrm{for}\:+\mathrm{ve}\:\mathrm{root}. \\ $$$$\mathrm{Hence}\:\sqrt{\mathrm{1}}\:=\:\mathrm{1}\:\mathrm{not}\:\sqrt{\mathrm{1}}=\pm\mathrm{1} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{write}\:\pm\sqrt{\mathrm{1}}=\pm\mathrm{1}\:\mathrm{but} \\ $$$$\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{write}\:\sqrt{\mathrm{1}}=\pm\mathrm{1} \\ $$

Commented by FilupSmith last updated on 19/Oct/16

but (√x^2 )=±x (+x)^2 =x^2 (−x)^2 =(−1)^2 (x)^2 =x^2

$$\mathrm{but}\:\sqrt{{x}^{\mathrm{2}} }=\pm{x} \\ $$$$\left(+{x}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$$\left(−{x}\right)^{\mathrm{2}} =\left(−\mathrm{1}\right)^{\mathrm{2}} \left({x}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$

Commented by nume1114 last updated on 19/Oct/16

(√x^2 )=∣x∣ example if x=3, (√x^2 )=(√9)=3,∣x∣=3 x=−1, (√x^2 )=(√1)=1,∣x∣=1 x=0, (√x^2 )=(√0)=0,∣x∣=0

$$\sqrt{{x}^{\mathrm{2}} }=\mid{x}\mid \\ $$$${example} \\ $$$${if}\:{x}=\mathrm{3},\:\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} }=\sqrt{\mathrm{9}}=\mathrm{3},\mid{x}\mid=\mathrm{3} \\ $$$$\:\:\:\:\:\:{x}=−\mathrm{1},\:\:\sqrt{{x}^{\mathrm{2}} }=\sqrt{\mathrm{1}}=\mathrm{1},\mid{x}\mid=\mathrm{1} \\ $$$$\:\:\:\:\:\:{x}=\mathrm{0},\:\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} }=\sqrt{\mathrm{0}}=\mathrm{0},\mid{x}\mid=\mathrm{0} \\ $$

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