Question

Determinati numerele reale a si b in fiecare dintre cazurile:
a)a²+4a+b²-6b+13=0
b)a²+b²+2(a+2b)+5=0
c)a²+2√3a+b²+4√2b+11=0
d)2a²+2√2a+3b²+2√3b+2=0
Va rog frumos! Am nevoie urgent!

Answer 1

$\displaystyle a^2+4a+b^2-6b+13=0 \\ a^2+4a+4+b^2-6b+9=0 \\ (a+2)^2+(b-3)^2=0 \\ x^2\geq 0\forall x\in R \\ \implies (a+2)^2\geq 0\forall a\in R \\ \implies (b-3)^2\geq 0\forall x\in R \\ Suma \ acestor \ patrate \ perfecte \ va \ fi \ 0 \ daca \ fiecare \ dintre \ ele \ va \ fi \ egal \ cu \ 0. \\ \implies (a+2)^2=0 \\ a+2=0 \\ a=-2 \\ \implies (b-3)^2=0 \\ b-3=0 \\ b=3 \\ \\ a^2+b^2+2(a+2b)+5=0 \\ a^2+2a+1+b^2+4b+4=0 \\ (a+1)^2+(b+2)^2=0 \\ (a+1)^2=0 \\ a+1=0 \\ a=-1 \\ (b+2)^2=0 \\ b+2=0 \\ b=-2 \\ \\ a^2+2\sqrt3a+b^2+4\sqrt2b+11=0 \\ a^2+2\sqrt3a+3+b^2+4\sqrt2b+8=0 \\ (a+\sqrt3)^2+(b+2\sqrt2)^2=0 \\ (a+\sqrt3)^2=0 \\ a+\sqrt3=0 \\ a=-\sqrt3 \\ (b+2\sqrt2)^2=0 \\ b+2\sqrt2=0 \\ b=-2\sqrt2 \\ \\ 2a^2+2\sqrt2a+3b^2+2\sqrt3b+2=0 \\ 2a^2+2\sqrt2a+1+3b^2+2\sqrt3b+1=0 \\ (\sqrt2a+1)^2+(\sqrt3b+1)^2=0 \\ (\sqrt2a+1)^2=0 \\ \sqrt2a+1=0 \\ a=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2} \\ (\sqrt3b+1)^2=0 \\ \sqrt3b+1=0 \\ b=-\frac{1}{\sqrt3}=-\frac{\sqrt3}{3}$