0.4

It doesn't matter. You still poo on the street.

Root of (3-2root2)=.423

this is just teacher porn. students learning at this level don’t need these special cases just designed to show off. Grade: D-

Here √2-1 should be open with mode.I know that it will open with +ve sign but still 😅 it will better if he first put the mode

-1

If b is = -1, then shouldn't -2(√2)(-1) equals to +2√2 then it'll be [(√2)^2 + (-1)^2 + (2√2)]

Can't root of 9 also be -3, and root 8 be –2 radical 2?

√2+1

You did not show that -2(square root of 2) times 1 = 2(square root of 2) times -1 such that b= -1

💀🥶

√2 – 1

No sería más fácil hacerlo con la calculadora

WHY ON EARTH, 1 = (-1)² ???!

looks good to me

It is quite correct, but only till class 10, in further classes we are introduced with the topic complex no.
So, there will be two different case, we can’t tell that whether root 2 is negative or 1 is negative…❤ so one of the equation will be imaginary and other will be real…

This is a very convoluted method with trial-and-error steps. I figured out that you can solve it in Q[sqrt(2)], treating sqrt(2) like an imaginary number (called z here):

sqrt(3-2z) where z^2 = 2; note solution must be >= 0

a+bz st (a+bz)^2 = 3-2z

(a+bz)^2 = a^2 + 2b^2 + 2abz

3 = a^2 + 2b^2, -2 = 2ab

-1 = ab => a = -b

3 = (-b)^2 + 2b = b^2 + 2b

=> b^2 + 2b – 3 = 0

=> b = -1 +- sqrt(1+3) = -1 +- 2 = -3 | 1

sqrt(3-2z) = 3 – 3z < 0 => reject

sqrt(3-2z) = -1 + z = sqrt(2) – 1 >= 0 => solution

3rd line …. Where did the -1 come from? Shouldn’t it be (1)^2?

Bhai 3-2root2 is also ok … if answer is allowed in root form why to calculate further

How is that solving it?