MGMAT Flash cards

[prachich1987]

The first few steps pf the problem are given
complete the steps & answer what would be the value of x?

sq.root(x+3)=x-3
x+3=(x-3)^2
x+3=x^2-6x+9
0=x^2-7x+6

[beat_gmat_09]

x^2 - 7x + 6
x can be 6 or 1
Any problem ?

[prachich1987]

x^2 - 7x + 6
x can be 6 or 1
Any problem ?

See below the explanation provided in MGMAT flash cards
I am not very convinced but

[beat_gmat_09]

x^2 - 7x + 6
x can be 6 or 1
Any problem ?

See below the explanation provided in MGMAT flash cards
I am not very convinced but

Sorry, i just focused on the quadratic equation part.
The explanation is correct. Plug back x = 1 in the stem
sqrt(x+3) = x - 3
=sqrt(1+3) = 1 - 3
=sqrt(4)=-2
2 ! = -2
(Editing: I hope you know this - sqrt(x^2) = x when x is +ve, but sqrt(x^2) = -x when x is -ve, try plugging values.)
x = 1 is not the solution as +2 != -2.

For x = 6
sqrt(6+3) = 6-3
sqrt(9) = 3
3 = 3
True, x = 6 is the solution.

[prachich1987]

quote

(I hope you know this - sqrt(x) = x when x is +ve, but sqrt(x) = -x when x is -ve, try plugging values.)
x = 1 does not fit because x + 3 is +ve sqrt (x+3) has to be +ve , not negative and as x - 3 is -ve x = 1 is not the solution..

unquote

sorry but i am not clear on the above part
how can square root of x will be x?
i consider it's a typo error
but square root of (2X2) will be +2 or -2.
do u mean to say square root of (2X2) is only 2?

[beat_gmat_09]

sorry but i am not clear on the above part
how can square root of x will be x?
i consider it's a typo error
but square root of (2X2) will be +2 or -2.
do u mean to say square root of (2X2) is only 2?

The detail i wanted to give was for something else, sorry for the confusion, but it is helpful anyways.

sqrt(4) = 2, square root of any number is +ve in GMAT (Please refer (1) ), but -
x^2 = 4 has different values of x, +2 and -2, (-2)^2 = 4 and (+2)^2 = 4 so x can take two values.

but Sqrt(4) is always +ve.

I forgot to put x^2 in the previous post
(1):
Correcting the part in previous post: sqrt(x^2) = x when x is +ve, but sqrt(x^2) = -x when x is -ve.
Check with +ve and -ve values.
x = -2 , x^2 = 4
sqrt(4) = 2 => 2 = -(x) i.e. 2 = -(-2) = 2
For x = +2 , x^2 = 4
sqrt(4) = 2 => 2 = x

The question above does not satisfy for x = 1, the reason is obviously +2 != -2
I hope you got me. Please let me know if you have any doubts, i don't want you to get confused because of my explanation.

Thanks.

[prachich1987]

sorry but i am not clear on the above part
how can square root of x will be x?
i consider it's a typo error
but square root of (2X2) will be +2 or -2.
do u mean to say square root of (2X2) is only 2?

The detail i wanted to give was for something else, sorry for the confusion, but it is helpful anyways.

sqrt(4) = 2, square root of any number is +ve in GMAT (Please refer (1) ), but -
x^2 = 4 has different values of x, +2 and -2, (-2)^2 = 4 and (+2)^2 = 4 so x can take two values.

but Sqrt(4) is always +ve.

I forgot to put x^2 in the previous post
(1):
Correcting the part in previous post: sqrt(x^2) = x when x is +ve, but sqrt(x^2) = -x when x is -ve.
Check with +ve and -ve values.
x = -2 , x^2 = 4
sqrt(4) = 2 => 2 = -(x) i.e. 2 = -(-2) = 2
For x = +2 , x^2 = 4
sqrt(4) = 2 => 2 = x

The question above does not satisfy for x = 1, the reason is obviously +2 != -2
I hope you got me. Please let me know if you have any doubts, i don't want you to get confused because of my explanation.

Thanks.

thanks for the above!
its clear now