How do you solve #5x ^ { 2} - 180= 0#?

2 Answers
seol
Dec 22, 2016

#\pm6\leftrightarrow x\leftrightarrow6#
See explanation for details

Explanation:

Original equation given
#5x^2-180=0#

Steps

  • (1) Add 180 to both sides and simplify
    #5x^2\cancel{\stackrel{+180}{-180}}=0+180#
    #5x^2=180#

  • (2) Divide both sides by 5 and simplify
    #(\cancel{5}x^2)/\cancel{5}=180/5#
    #x^2=36#

  • (3) Square root both sides and simplify
    #\stackrel{x}{\cancel(\sqrt(x^2))}=\sqrt(36)#
    #x=\sqrt(36)=\pm6#

Usually, a square root has a negative and a positive result, as #(-x)^2=x^2# Depending on how many answers/results your question wants, your solution can be #\pm6# or just #6#.

Jim G.
Dec 22, 2016

#x=+-6#

Explanation:

#"Isolate " 5x^2# by adding 180 to both sides of the equation.

#5x^2cancel(-180)cancel(+180)=0+180#

#rArr5x^2=180#

divide both sides by 5

#(cancel(5) x^2)/cancel(5)=180/5#

#rArrx^2=36#

To solve for x, take the #color(blue)"square root of both sides"#

#sqrt(x^2)=+-sqrt36#

#rArrx=+-6" are the solutions"#

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