Given that #sqrt(x^2+4)=-3x# determine the value or values of #x#?

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Answer 1 · 2016-06-27T13:34:08

#x=+-sqrt(2)/2#

The x-intercept is at #y=0#

Write as:#" "0=3x-sqrt(x^2+4)#

Add #sqrt(x^2+4)" "# to both sides

#sqrt(x^2+4)=3x#

Square both sides

#x^2+4=9x^2#

Subtract#" "x^2# from both sides

#8x^2=4#

Divide both sides by 8

#x^2=1/2#

Square root both sides

#x=+-1/sqrt(2)#
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Mathematicians consider it bad practice to have roots in the denominator if at all avoidable.

Multiply b1 but in the form of #1=sqrt(2)/ sqrt(2)#

#x=+-1/sqrt(2)xxsqrt(2)/ sqrt(2) = +-sqrt(2)/2#