Question

x²+6x+24=10*sqrt{ x^{2} +6x}

Answer 1

$x^{2} +6x geq 0 \ x^{2} +6x = 0 \ x(x+6)=0 \ x=0;x=-6$
x∈(-oo;-6] U [0;+oo)
$x^{2} +6x=t \ x^{2} +6x+24=10*sqrt{ x^{2} +6x} \ t+24=10 sqrt{t} \ (t+24) ^{2} =(10 sqrt{t}) ^{2} \t ^{2} +48t+576=100t\ t ^{2} -52t+576=0 \ D=52 ^{2} -4*576=2704-2304=400=20^{2} \ t= frac{52-20}{2} =16 \ t= frac{52+20}{2} =36 \ t=x^{2} +6x \ x^{2} +6x=16 \ x^{2} +6x-16 =0\ x=-8 \ x=2 \ x^{2} +6x=36 \ x^{2} +6x-36=0\ D=36-4*(-36)=36+144=180=(6 sqrt{5} ) ^{2} \ x= frac{-6+6 sqrt{5} }{2} =-3+3 sqrt{5} \ x=-3-3 sqrt{5}$