Solve. 5x2 – 20 = 0 x = 5 and x = 4 x = 4 and x = 2 x = –5 and x = 20 x = 2 and x = –2
Step 1: Add 20 on both sides
Step 2: divide both sides by 5
Can you try that?
Sorry, did i make any mistake? :S
If you factored out 5 first then you will get: \[5(x^2 - 4) = 0 \implies 5(x^2 - 2^2) = 0\]
IM CONFUSED
\(5x^2-20=0 \) \(x^2 = 4 \) When you take square root on both sides you'll get two solutions, right?
yes but sometimes it will not be applicable.. Like: \[x^2 = 4x\] you can't cancel x here..
You are right diya sorry my misunderstanding...
Its Okay, No Problem :)
@harrisonshields just do the same what diya told you to do..
ok so if x^2 = 4
@harrisonshields \(5x^2-20=0 \) Add 20 on both sides \(5x^2 - 20 +20 = 20 \) \(5x^2=20\) Divide both sides by 5 \(5x^2/5 = 20/5 \) \(x^2 = 4 \)
i get 2 and 4/2?
Don't you think they both are same @harrisonshields
oh so - and + 2
\[\sqrt{x^2}=\sqrt{4}\]\[x= \pm 2\]
Are you hungry?? Then why you ate 2?? @harrisonshields ha ha ha.
Right! @harrisonshields \(\pm 2 \) because \( 2^2 = (-2)^2 = 4\)
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