vedha0 wrote:
Bunuel bb EMPOWERgmatRichC
can you please clarify thiscase?
Kanna444 wrote:
x^3 = 8 actually has 3 solutions;
\( x^3 - 8=0\)
\((x-2)(x^2+2x+4)=0\)
1 for the above equation\(x=2\)
other 2 solutions are imaginary\( [-1+i\sqrt{3} ],[-1-i\sqrt{3}]\)
so we cant say x is an interger or imaginary,
can someone pleaseclarify.
Bunuel
EMPOWERgmatRichC
\( x^3 - 8=0\)
\((x-2)(x^2+2x+4)=0\)
1 for the above equation\(x=2\)
other 2 solutions are imaginary\( [-1+i\sqrt{3} ],[-1-i\sqrt{3}]\)
so we cant say x is an interger or imaginary,
can someone pleaseclarify.
Bunuel
EMPOWERgmatRichC
On the GMAT, all numbers are assumed to be real numbers by default.
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Statistics : Posted by Bunuel • on 26 Apr 2019, 02:16 • Replies 15 • Views 14262
