Question

Rezolvati in R ecuatia:
a)4x²-25=0
b)$\frac{1}{4}z^{2}-16=0$

Answer 1

$a) 4x^{2}-25=0 <=> (2x-5)(2x+5)=0 =>$ 2x-5=0 sau 2x+5=0.
$2x-5=0 => 2x=5 => x= \frac{5}{2} \\ 2x+5=0 => 2x=-5 => x=- \frac{5}{2}$
Solutie: {$- \frac{5}{2}$;$\frac{5}{2}$}

$b) \frac{1}{4} z^{2} -16=0 <=> ( \frac{z}{2}-4)( \frac{z}{2}+4)=0 =>$ $\frac{z}{2}-4=0$ sau $\frac{z}{2}+4=0$.
$\frac{z}{2}-4=0 => \frac{z}{2}=4 => z=8 \\ \frac{z}{2}+4=0 => \frac{z}{2}=-4 => z=-8$
Solutie: {-8;8}.

*Am aplicat formula $x^{2} - y^{2} =(x+y)(x-y)$.

Answer 2

a.  (2x-5)(2x+5)=0
     2x-5=0  x=2.5           
     2x+5=0  x=-2.5      
b    1/4z²-16=0
      (1/2z-4)(1/2z+4)=0
       1/2z-4=0
        z=8
       1/2z+4=0
         z=-8