Controversial Question
Find the value of x satisfying
x + 1 = 0 .
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8 solutions
Moderator note:
Great solution. They may not always exist a root to an equation.
BY squaring on both sides we get x = 1. What u say abt that????
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hi Samara,actually when you square an equation it is no longer the same equation.for example ,X^2= -1, has no real roots but if u square this eqn u have X^4=1,which has two real solution ,+1,-1:because when you square this equation ,in addition with given equation it also contains an extra equation [x^2 - 1], which gives two real roots.
If x = 1 , then x + 1 = 1 + 1 = 2 = 0 . It does not satisfy the given equation and its kinda an extraneous solution
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cant we write sqrt(1) = -1???
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@Yash Mehan – Nope , square root is defined as x 2 = ∣ x ∣ , hence it always gives positive solution.
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@Nihar Mahajan – Nope Square root of any number is defined as plus or minus the number.
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@Samara Simha Reddy – Its a common misconception . Click here and read the wiki.
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@Nihar Mahajan – Must be a US thing. In the UK, we always accept that a square root has 2 values, but by convention take the positive.
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@Paul Simpson – So do we. I'm not sure what the difference is.
(By the way, your profile says you're from Indianapolis, in the US. Just thought I'd point that out.)
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@Whitney Clark – I am now! I used to teach Maths in the UK.....
Then x might be an imaginary number.
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@Samara Simha Reddy – How about i^4
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@Quang Trung Nguyễn – Nope, i 4 = 1 = 1 since i 4 = 1 .
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@Paul Ryan Longhas – how about (i^4)^1/2 = i^2 = -1
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@Joener Vidal – ( i 4 ) 2 1 = 1 2 1 = 1 since i 4 = 1 . Clearly, not all the time ( a ) b = a b is true.
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@Paul Ryan Longhas – i see. thank you @Paul Ryan Longhas
@Paul Ryan Longhas – How do you know this time it works and this time it doesn't.
@Joener Vidal – Your solution has an error. Well to make it clear, use the concept that a 2 = ∣ a ∣ . In your solution, i 4 = ∣ i 2 ∣ = ∣ − 1 ∣ = 1 .
Hi Nihar: The question is which value can satisfy the equation so x=1 can satisfy the equation because root of 1 is +-1. So you can consider -1 for this equation.
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@Aman Parauliya – Absolutely!
@Aman Parauliya – No, the square root symbol refers to the principal root only.
even so, re-substituting in the original equation you get 2=0, hence still a contradiction. Edit: I didn't read the other posts, apologies for repeating what others already said
x + 1 = 0 ⇒ x + 1 = 0 ⇒ x = − 1 ⇒ ( x ) 2 = ( − 1 ) 2 ⇒ x = 1
Why can't this be??
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Consider the graph of the function y = x , we see that the graph of this function is not the same in the graph of the function y 2 = x . So, squaring both side is not always allowed to use especially if we dealing in negative numbers. Clearly, if we raised the equation in a , the solution gives an extraneous root/s or sometimes the solution is not completed find.
Because the square root symbol refers to the principal root only and squaring introduced extraneous solutions.
sqrt(x)+1 = 0 --> sqrt(x)=-1 --> x=1
in fact: if x=1 --> sqrt(x)=+- 1, and sqrt(x)+1=0 is a possibility
@Nihar Mahajan nice question. Just needed to know why the answer can't be a complex number since you asked for ANY solution possible.
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No complex number has -1 for a principal square root.
√x + 1 = 0 |||| √x = -1 ||| (√x)^2 = (-1)^2 ||| x = 1 |||
√1 = +1 / -1
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Wrong, the square root symbol refers to the principal root only, not to both.
I got the write answer because no other solution was right but is the value of x was i 2 , then then equation would have satisfied So, the option, no solution exists is technically wrong.
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x = i 2 does not satisfy the equation. Check again. No solution exists is technically correct.
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x
+
1
=
0
.
substituting x =
i
2
i
2
+
1
=
0
.
+
1
=
0
−
1
+
1
=
0
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@Ashish Menon – i 2 + 1 = i + 1 = 0
Or do you think i = − 1 ?
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@Nihar Mahajan – ok got it, sorry for the inconvenience
@Nihar Mahajan – what if i put x= i^4; in that case sqrt (i^4) = i^2 = -1.
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@Manish Kumar Singh – i 4 = 1 . So, square of 1 is 1
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@Ashish Menon – no brother,what if i will 1st take square root of i^4 and then put in to equation. sqrt( i^4 )=i^2= -1.and it satisfies the given equation.
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@Manish Kumar Singh – Use the fact that x 2 = ∣ x ∣ . Thus, i 4 = ∣ i 2 ∣ = ∣ − 1 ∣ = 1 .
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@Paul Ryan Longhas – agree,but i want to know that whether these rules are valid in case we deal with imaginary numbers.
Nihar, we know that 6 4 = ± 8
( − 8 ) 2 = 6 4
so 6 4 has two values: + 8 a n d − 8
so 1 can also be equal to + 1 or − 1
so if x = − 1 , so x could have taken value of 1 .......
pls tell me where i am wrong, coz i still feel my concepts shaken up in squares and square roots.
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6 4 = − 8 . Remember that x 2 = ∣ x ∣ and not x . This is a common misconception
what about "i^4"?
-1 times -1 = 1 there fore the root of 1 =1, -1 so -1 + 1 = 0
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Sorry, no. "The" root is nonnegative; "a" root can be anything that fits.
"By convention", I guess a lot of professional mathematicians would get this wrong. In any case, this is a poor question as it doesn't test reasoning in any way.
x + 1 = 0 Now x = − 1
But a square root can't give negative value,so no solution exists
By definition i^2=-1 so \sqrt(-1) can be either i or -i. It's not correct that no solution exists.
√1 has two solutions +1 & -1. One of the solutions can satisfy the equation (i.e -1).
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nope, the symbol is use for taking principal square root.
exactly. x could have been 1. right?
What if x=i^4 wouldn't the square root then be i^2 which would be -1?
Moderator note:
No, the square root of i 4 = 1 is just 1.
sqrt(x) has to equal -1, but that does not exist, and therefore the equation has no solutions
Précisez dans quel ensemble résoudre cette équation : IR ou ℂ
Square root of any number will get positive number , it is impossible to get A negative number.
x is non negative and not equal to -1 for all x.
True, @Nihar Mahajan , thank you for making me understand it
why is it , that the value x isn't negative two?
x + 1 = 0 ⇒ x = − 1
By convention , taking square root always gives a positive solution whereas above its giving a negative value. Hence , contradiction and no solution possible.