Question

Determinați soluțiile ecuațiilor:
a) (2x+3)²+7=4(x-2)²
b) (3x-4)²+20=5(x+1)²+(2x-3)²
c) (x+2)(3x+2)-3x=2x²+(x+1)²

Answer 1

[tex]a)(2x+3)^2+7=4*(x-2)^2\\ 4x^2+12x+9+7=4*(x^2-4x+4)\\ 4x^2+12x+16=4x^2-16x+16\\ 4x^2-4x^2+12x+16x=16-16\\ 0+28x=0\\ x=28:0\\ \boxed{x=0}\\\\\\ b)(3x-4)^2+20=5*(x+1)^2+(2x-3)^2\\ 9x^2-24x-16+20=5*(x^2+2x+1)+4x^2-12x+9\\ 9x^2-24x+4=5x^2+10x+5+4x^2-12x+9\\ 9x^2-5x^2-4x^2-24x-10x+12x=5+9-4\\ -22x=10\\ x= \frac{-22}{10}\\ \boxed{x=-2,2} [/tex][tex]c)(x+2)(3x+2)-3x=2x^2+(x+1)^2\\ 3x^2+2x+6x+4-3x=3x^2+x^2+2x+1\\ 3x^2-3x^2-x^2+2x+6x-3x-2x=1-4\\ -x^2-3x=-3===\ \textgreater \ \ \textgreater \ ecuatia~ nu~ are ~solutie!!![/tex]