Question

Rezolvati urmatorul exercitiu: Va multumesc! <3 <3  :) :D ^.^
a) 2x² - 5x + 2 = 0
b) 5x²+ 14x - 3 =0
c) x² - 4x +3 = 0
d) x² + 10x + 24 = 0
e) x² - 4x - 21 = 0
f) x² - 4x + 1 = 0
g) x² + 2x - 4 = 0

Answer 1

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

$\displaystyle b).5x^2+14x-3=0 \\ a=5,b=14,c=-3 \\ D=b^2-4ac=14^2-4 \cdot 5 \cdot (-3)=196+60=256 \\ x_1= \frac{-14+ \sqrt{256} }{2 \cdot 5} = \frac{-14+16}{10} = \frac{2}{10} = \frac{1}{5} \\ \\ x_2= \frac{-14- \sqrt{256} }{2 \cdot 5} = \frac{-14-16}{10} = \frac{-30}{10} =-3$

$\displaystyle c).x^2-4x+3=0 \\ a=1,b=-4,c=3 \\ D=b^2-4ac=4^2-4 \cdot 1 \cdot 3=16-12=4\ \textgreater \ 0 \\ x_1= \frac{4+ \sqrt{4} }{2 \cdot 1} = \frac{4+2}{2} = \frac{6}{2} =3 \\ \\ x_2= \frac{4- \sqrt{4} }{2 \cdot 1} = \frac{4-2}{2} = \frac{2}{2} =1$

$\displaystyle d).x^2+10x+24=0 \\ a=1,b=10,c=24 \\ D=b^2-4ac=10^2-4 \cdot 1 \cdot 24=100-96=4\ \textgreater \ 0 \\ x_1= \frac{-10+ \sqrt{4} }{2 \cdot 1} = \frac{-10+2}{2} = \frac{-8}{2} =-4 \\ \\ x_2= \frac{-10- \sqrt{4} }{2 \cdot 1} = \frac{-10-2}{2} =- \frac{12}{2} -6$

$\displaystyle e).x^2-4x-21=0 \\ a=1,b=-4,c=-21 \\ D=b^2-4ac=4^2-4 \cdot 1 \cdot (-21)=16+84=100\ \textgreater \ 0 \\ x_1= \frac{4+ \sqrt{100} }{2 \cdot 1} = \frac{4+10}{2} = \frac{14}{2} =7 \\ \\ x_2= \frac{4- \sqrt{100} }{2 \cdot 1} = \frac{4-10}{2} = \frac{-6}{2} =-3$

$\displaystyle f).x^2-4x+1=0 \\ a=1,b=-4,c=1 \\ D=b^2-4ac=4^2-4 \cdot 1 \cdot 1=16-4=12\ \textgreater \ 0 \\ x_1= \frac{4+ \sqrt{12} }{2 \cdot 1} = \frac{4+2 \sqrt{3} }{2} = \frac{2(2+ \sqrt{3} )}{2} =2+ \sqrt{3} \\ \\ x_2= \frac{4- \sqrt{12} }{2 \cdot 1} = \frac{4-2 \sqrt{3} }{2} = \frac{2(2- \sqrt{3} )}{2} =2- \sqrt{3}$

$\displaystyle g).x^2+2x-4=0 \\ a=1,b=2,c=-4 \\ D=b^2-4ac=2^2-4 \cdot 1 \cdot (-4)=4+16=20\ \textgreater \ 0 \\ x_1= \frac{-2+ \sqrt{20} }{2 \cdot 1} = \frac{-2+2 \sqrt{5} }{2} = \frac{2(-1+ \sqrt{5} )}{2} =-1+ \sqrt{5} \\ \\ x_2= \frac{-2- \sqrt{20} }{2 \cdot 1} = \frac{-2-2 \sqrt{5} }{2} = \frac{2(-1- \sqrt{5} )}{2} =-1- \sqrt{5}$