Question

Rezolvati ecuatia: (x+3)(x+5)++(1-x)la patrat =2(3-x)+(2x-1)la patrat +7 la patrat e doar paranteza dinaintea cuvantului
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Answer 1

$(x+3)(x+5)+(1-x)^2=2(3-x)+(2x-1)^2+7 \\ x^2+5x+3x+15+1-2x+x^2=6-2x+4x^2-4x+1+7 \\ x^2+x^2-4x^2+5x+3x-2x+2x+4x=6+1+7-15-1 \\ -2x^2+12x=-2 \\ -2x^2+12x+2=0 \\ a=-2,b=12,c=2 \\ D= b^2-4ac=12^2-4*(-2)*2=144+16=160\ \textgreater \ 0 \\ x_1= \frac{-12+ \sqrt{160} }{2*(-2)} = \frac{-12+4 \sqrt{10} }{-4} = -(-3+ \sqrt{10} )=3- \sqrt{10} \\ x_2= \frac{-12- \sqrt{160} }{2*(-2)} = \frac{-12-4 \sqrt{10} }{-4} =-(-3- \sqrt{10} )=3+ \sqrt{10} \\ S= \left \{ 3- \sqrt{10} ;3+ \sqrt{10} \left \}$