Question

Am nevoie sa ma ajutați la exercițiul 4. Mulțumesc

Answer 1

sper sa te ajute răspunsul meu!

Answer 2

$\displaystyle \it a).E(x)=(2x+1)^2-(x-1)^2+2(x-3)(x+3)-x(4x+3)+17 \\ E(x)=(2x)^2+2 \cdot 2x \cdot 1+1^2-(x-1)^2+2(x^2-9)-4x^2-3x+17 \\ E(x)=4x^2+4x+1-(x^2-2 \cdot x \cdot 1+1^2)+2x^2-18-4x^2-3x+17 \\ E(x)=4x^2+4x+1-(x^2-2x+1)+2x^2-18-4x^2-3x+17 \\ E(x)=4x^2+4x+1-x^2+2x-1+2x^2-18-4x^2-3x+17 \\ E(x)=x^2+3x-1 \\ b).E(x)=3 \\ x^2+3x-1=3 \\ x^2+3x-1-3=0 \\ x^2+3x-4=0 \\ a=1,~b=3,~c=-4 \\ \Delta=b^2-4ac=3^2-4 \cdot 1\cdot (-4)=9+16=25\ \textgreater \ 0 \\ x_1= \frac{-3+ \sqrt{25} }{2 \cdot 1} = \frac{-3+5}{2} = \frac{2}{2} =1$
$\displaystyle \it x_2= \frac{-3- \sqrt{25} }{2 \cdot 1} = \frac{-3-5}{2} = \frac{-8}{2} =-4$