Yuii
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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

What are the solutions to the equation?

5x3 = 405x

PLEASE HELP ASAP CORRECT ANSWER ONLY PLEASE What are the solutions to the equation 5x3 405x class=

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gmany

[tex]5x^3=405x\ \ \ \ |-405x\\\\5x^3-405x=0\\\\5x(x^2-81)=0\iff 5x=0\ \vee\ x^2-81=0\\\\5x=0\ \ \ \ |:5\\\boxed{x=0}\\\\x^2-81=0\ \ \ |+81\\x^2=81\to x=\pm\sqrt{81}\to \boxed{x=\pm9}\\\\Answer:\ x=-9\ or\ x=0\ or\ x=9[/tex]

highwireact

There's one housekeeping item we can take care of first:

  • 5x³ and 405x have the common factor 5x [5x³ = (5x)(x²) and 405x = (5x)(81)]. We can divide both side of the equation without changing the equality, obtaining the result [tex]x^2=81[/tex]

What this new equality is asking is: "What number times itself gives us 81?" There are actually two answers to this question: 9 and -9 (since multiplying a negative number by itself makes it positive).

Our solution then is [tex]x=\pm9[/tex]

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