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
Nu stiu sa fac semnul din tastatura!Va ros sa ma ajutati!E urgent!!!

Answer 1

(x+3)(x+5)+(1-x)²=2(3-x)+(2x-1)²+7²
x²+5x+3x+15+1-2x+x²=6-2x+4x²-4x+1+7
2x²+6x+16=4x²-6x+14
2x²-4x²+6x+6x+16-14=0
-2x²+12x+2=0
2(-x²+6x+1)=0
-x²+6x+1=0             inmultim -1
x²-6x-1=0
Δ=b²-4ac
a=1
b=-6
c=-1
Δ=36+4=40
x₁=(6-√40)/2=(6-2√10)/2=2(3-√10)/2=3-√10
x₂=(6+√40)/2=(6+2√10)/2=2(3+√10)/2=3+√10
D={3-√10, 3+√10}

Answer 2

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