Question 1172104
let x^2+3x=w^2
so w^2+w-6=0
(w+3)(w-2)=0
w=-3, w=2, but w=-3 is extraneous, as sqrt (x^2+3x) cannot be negative.
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(x^2+3x-4=0
(x+4)(x-1)=0
x=-4, 1
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{{{graph(300,300,-10,10,-10,10,x^2+3x+(sqrt(x^2+3x))-6)}}}