How do you solve #4x^2=25#?

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Answer 1 · 2016-04-17T04:55:09

#x=+-5/2#

Use the #color(gold)("golden rule"# of Algebra

#"What we do on one side,must be done on the other side also"#

Our aim is to isolate #x# to find its value

#color(blue)(4x^2=25#

Divide both sides by #4#

#rarr(cancel4x^2)/cancel4=25/4#

#rarrx^2=25/4#

Take the square root of both sides

#rarrsqrt(x^2)=sqrt(25/4#

#rarrx=sqrt(25/4)#

Remember that,if we take the square root of a number,it can be positive or negative

So,

#rarrx=+-sqrt(25/4)#

Use the property

#color(brown)(sqrt(x/y)=sqrtx/sqrty#

#rarrx=+-sqrt25/sqrt4#

#color(blue)(rArrx=+-5/2#

Remember that the symbol #+-# means Plus or Minus

it indicates that

#rArrcolor(blue)(x=5/2,-5/2#