math solving questions

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bnjnkn
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#1
Report Thread starter 5 months ago
#1
the question is to solve 2x+√x-6 using substitution .

Why is it that you only get one solution ?
Let me explain I substituted x with v^2. Once I got an equation which i factorized to (2v-3)(v+2) ,i solved for v which is v=3/2 and v=-2. According to my substitution, u have square v to get x ,(x=v^2). So far i am right .however why is the mark scheme is shown only to square 3/2 to get x , even though u could also square -2?what i sthe mathematic logic behind it ? ah i also subsituted 4 but the orginal equation did not equal to 0.so does it mean that i always to check in these type of equation or is there a reasoning behind it ? oh yhh why is v positive
Last edited by bnjnkn; 5 months ago
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ashvinsingh
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#2
Report 5 months ago
#2
(Original post by magnetodespacito)
if ur substituting x with v^2 then x=v^2, so i think u would have to square root ur solutions to get x. the solutions for v^2 are 3/2 and -2, but there's only one solution for x as you can't root -2. (cant root negatives)
what does the mark scheme say the answer is ?
It’d be the opposite way around. X = v^2 so the values of X would be 9/4 and 4. (V=3/2, V=-2)
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bnjnkn
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#3
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#3
(Original post by ashvinsingh)
It’d be the opposite way around. X = v^2 so the values of X would be 9/4 and 4. (V=3/2, V=-2)
but i know the answer if u substitute into the equation,it does not give a zero
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bnjnkn
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#4
Report Thread starter 4 months ago
#4
(Original post by ashvinsingh)
Oh okay. As you are replacing X with V^2, X must be positive. As any number squared >=0 as X is positive the equation 2x +root x -6 =0 only has one solution. If X was negative you wouldn’t be able to use the square root. So only one solution.
but 4 is positive
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mqb2766
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#5
Report 4 months ago
#5
(Original post by bnjnkn)
but 4 is positive
x = v^2
sqrt(x) = v

So what is the value of x for v=3/2 and v=-2? Note the sqrt() means take the positive root, so the root in
2x+√x-6
is always positive i.e. √4 = 2. It does not equal -2.
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bnjnkn
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#6
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#6
(Original post by mqb2766)
x = v^2
sqrt(x) = v

So what is the value of x for v=3/2 and v=-2? Note the sqrt() means take the positive root, so the root in
2x+√x-6
is always positive i.e. √4 = 2. It does not equal -2.
why does sqrt means to take positive. here to convert from v to x you are not rooting anything,you are squaring it,for exampkle from x to v,u square . if u sqaure both numbers 3/2 and -2,u get a positive number which can be square root again to get same numbers. should not the rule >root 4 equals to negative2 and positive 4 apply?
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mqb2766
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#7
Report 4 months ago
#7
(Original post by bnjnkn)
why does sqrt means to take positive. here to convert from v to x you are not rooting anything,you are squaring it,for exampkle from x to v,u square . if u sqaure both numbers 3/2 and -2,u get a positive number which can be square root again to get same numbers. should not the rule >root 4 equals to negative2 and positive 4 apply?
There are two solutions (+/-2) to the equation
x^2 = 4
However, there is just a single value (+2) for
x = sqrt(4)

sqrt() is a function which returns a single value, the positive root. If you want the negative value, you'd have to explicitly write
x = -sqrt(4)
like in the quadratic formula where you have +/-sqrt(b^2-4ac) so show that you want both the postive and the negative values.

In the equation you have sqrt(x), this is the positive value.
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bnjnkn
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#8
Report Thread starter 4 months ago
#8
(Original post by mqb2766)
There are two solutions (+/-2) to the equation
x^2 = 4
However, there is just a single value (+2) for
x = sqrt(4)

sqrt() is a function which returns a single value, the positive root. If you want the negative value, you'd have to explicitly write
x = -sqrt(4)
like in the quadratic formula where you have +/-sqrt(b^2-4ac) so show that you want both the postive and the negative values.

In the equation you have sqrt(x), this is the positive value.
never knew that ,honestly ,our school been teaching wrong things like sqrt can be positive and negative .Thank u
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mqb2766
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#9
Report 4 months ago
#9
(Original post by bnjnkn)
never knew that ,honestly ,our school been teaching wrong things like sqrt can be positive and negative .Thank u
A bit more discussion here
https://math.stackexchange.com/quest...alued-function
but you should think of sqrt() as the inverse of ^2 so that must be one-to-one. To make the square function invertible, you restrict the domain to non-negative values [0,inf), hence that is the range of the inverse function sqrt().

It should be noted that when you wrote x = v^2, you introduced extraneous solutions into the equation
https://en.wikipedia.org/wiki/Extran...sing_solutions
So if youre unsure about the set of final solutions, sub them back into the original equation to verify.
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