math solving questions

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)

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

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?

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