# math solving questions

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bnjnkn

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#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

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

(Original post by

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 ?

**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 ?

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bnjnkn

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#3

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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)

**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)

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bnjnkn

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#4

(Original post by

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.

**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.

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mqb2766

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#5

(Original post by

but 4 is positive

**bnjnkn**)but 4 is positive

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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#6

(Original post by

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.

**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.

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mqb2766

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#7

(Original post by

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?

**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?

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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#8

(Original post by

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.

**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.

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mqb2766

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#9

(Original post by

never knew that ,honestly ,our school been teaching wrong things like sqrt can be positive and negative .Thank u

**bnjnkn**)never knew that ,honestly ,our school been teaching wrong things like sqrt can be positive and negative .Thank u

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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