Question

Ce obtinem dezvoltand binomul : (2x-5) totu la puterea a doua

Answer 1

[tex](2x-5)^2\\\\ --------------------\\ aplicam~formula~(a-b)^2=a^2-2ab+b^2\\ In~acest~caz,a=2x~si~b=5\\\\ ---------------------\\ =\ \textgreater \ (2x)^2-2*2x*5+5^2\\ =\ \textgreater \ 4x^2-20x+25\\\\ ---------------\\ Lasam~asa~deoarece~nu~avem~aceasi~termeni...[/tex]

Answer 2

(x+2)^2=(2x-5)^2 x^2+4x+4=4x^2-20x+25 (x^2-4x+4)-(4x^2-20x+25)=0 -3x^2+24x-21=0 -3*(x^2+8x-7)=0 x^2+8x-7=0:-3 x^2+8x-7=0 Coeficienti: A=1 B=-8 C=7 x=[8+/-√(-8^2-4*1*7)]:2*1 x=[8+/-√(64-28)]:2 x=(8+/-√36):2 x=(8+/-6):2 x1=(8+6):2 x1=14:2 x1=7 x2=(8-6):2 x2=2:2 x2=1 Solutia 1: x=7 Solutia 2: x=1