Question

Efectuati:5(x-3)^2-(2x-3)^2+x+3=4(x-2)^2-3(1-x)^2+5

Answer 1

[tex]5(x-3)^2-(2x-3)^2+x+3=4(x-2)^2-3(1-x)^2+5\\ 5(x^2-6x+9)-4x^2+12x-9+x+3=4(x^2-4x+4)-\\ -3(1-2x+x^2)+5\\ 5x^2-30x+45-4x^2+13x-6=4x^2-16x+16-3+6x-3x^2+5\\ x^2-17x+39=x^2-10x+18\\ -7x=-21\Rightarrow x=3[/tex]

Answer 2

$\displaystyle 5(x-3)^2-(2x-3)^2+x+3=4(x-2)^2-3(1-x)^2+5 \\ 5(x^2-2 \cdot x \cdot 3+3^2)-[(2x)^2-2 \cdot 2x \cdot 3+3^2]+x+3= \\ =4(x^2-2 \cdot x \cdot 2+2^2)-3(1^2-2 \cdot 1 \cdot x+x^2)+5 \\ 5(x^2-6x+9)-(4x^2-12x+9)+x+3=4(x^2-4x+4)- \\ -3(1-2x+x^2)+5 \\ 5x^2-30x+45-4x^2+12x-9+x+3=4x^2-16x+16-3+6x- \\ -3x^2+5 \\ 5x^2-4x^2-4x^2+3x^2-30x+12x+x+16x-6x=16-3-45+9- \\ -3+5 \\ -7x=-21 \Rightarrow x= \frac{-21}{-7} \Rightarrow x=3$