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Mathematics 74 Online
OpenStudy (anonymous):

Solve. 5x2 – 20 = 0 x = 5 and x = 4 x = 4 and x = 2 x = –5 and x = 20 x = 2 and x = –2

OpenStudy (diyadiya):

Step 1: Add 20 on both sides

OpenStudy (diyadiya):

Step 2: divide both sides by 5

OpenStudy (diyadiya):

Can you try that?

OpenStudy (diyadiya):

Sorry, did i make any mistake? :S

OpenStudy (anonymous):

If you factored out 5 first then you will get: \[5(x^2 - 4) = 0 \implies 5(x^2 - 2^2) = 0\]

OpenStudy (anonymous):

IM CONFUSED

OpenStudy (diyadiya):

\(5x^2-20=0 \) \(x^2 = 4 \) When you take square root on both sides you'll get two solutions, right?

OpenStudy (anonymous):

yes but sometimes it will not be applicable.. Like: \[x^2 = 4x\] you can't cancel x here..

OpenStudy (anonymous):

You are right diya sorry my misunderstanding...

OpenStudy (diyadiya):

Its Okay, No Problem :)

OpenStudy (anonymous):

@harrisonshields just do the same what diya told you to do..

OpenStudy (anonymous):

ok so if x^2 = 4

OpenStudy (diyadiya):

@harrisonshields \(5x^2-20=0 \) Add 20 on both sides \(5x^2 - 20 +20 = 20 \) \(5x^2=20\) Divide both sides by 5 \(5x^2/5 = 20/5 \) \(x^2 = 4 \)

OpenStudy (anonymous):

i get 2 and 4/2?

OpenStudy (anonymous):

Don't you think they both are same @harrisonshields

OpenStudy (anonymous):

oh so - and + 2

OpenStudy (diyadiya):

\[\sqrt{x^2}=\sqrt{4}\]\[x= \pm 2\]

OpenStudy (anonymous):

Are you hungry?? Then why you ate 2?? @harrisonshields ha ha ha.

OpenStudy (diyadiya):

Right! @harrisonshields \(\pm 2 \) because \( 2^2 = (-2)^2 = 4\)

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