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Mathematics 7 Online
OpenStudy (anonymous):

solve2x^2 =32

OpenStudy (asnaseer):

first divide both sides by 2 - what do you get?

OpenStudy (anonymous):

32 divided by 2 is 16

OpenStudy (asnaseer):

thats the right-hand-side, what about the left-hand-side of the equation?

OpenStudy (anonymous):

2 divided by 2 = 1 right

OpenStudy (asnaseer):

yes - so what is the resulting equation after dividing both sides by 2?

OpenStudy (anonymous):

2^2 =16 right ?

OpenStudy (asnaseer):

not quite, you started with:\[2x^2=32\]where has the \(x^2\) gone?

OpenStudy (anonymous):

oh yes umm we didit solve forx s it would be x^2 = 16 ?

OpenStudy (asnaseer):

perfect! no take the square roots of both sides and what do you get?

OpenStudy (asnaseer):

*now

OpenStudy (anonymous):

x = 256 because 16x16 = 256

OpenStudy (asnaseer):

not quite - you have squared 16 instead of taking its square root. what number, when multiplied by itself, will give you 16?

OpenStudy (anonymous):

4

OpenStudy (asnaseer):

thats it - well done. so, to summarise the steps, you did:\[2x^2=32\]then divide both sides by 2 to get:\[x^2=16\]then take square roots of both sides to get:\[x=4\]

OpenStudy (asnaseer):

have been taught that the square root can be negative as well?

OpenStudy (asnaseer):

*have you been...

OpenStudy (anonymous):

yes

OpenStudy (asnaseer):

ok, in that case the answer should really be:\[x=\pm4\]

OpenStudy (asnaseer):

hope that makes sense?

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