Question

Va rog ajutor la aceste ecuatii

Hellpppp

Answer 1

$\displaystyle a)3x^2-7x+4=0 \\ \\ a=3,~b=-7,~c=4\\ \\ \Delta=b^2-4ac=(-7)^2-4 \cdot 3 \cdot 4=49-48=1\ \textgreater \ 0 \\ \\ x_1= \frac{-(-7)- \sqrt{1} }{2 \cdot 3} = \frac{7-1}{6} = \frac{6}{6} =1 \\ \\ x_2= \frac{-(-7)+ \sqrt{1} }{2 \cdot 3} = \frac{7+1}{6} = \frac{8}{6} = \frac{4}{3}$
$\displaystyle b)5x^2-8x+3=0 \\ \\ a=5,~b=-8,~c=3 \\ \\ \Delta=b^2-4ac=(-8)^2-4 \cdot 5 \cdot 3=64-60=4\ \textgreater \ 0 \\ \\ x_1= \frac{-(-8)- \sqrt{4} }{2 \cdot 5} = \frac{8-2}{10} = \frac{6}{10} = \frac{3}{5} \\ \\ x_2= \frac{-(-8)+ \sqrt{4} }{2 \cdot 5} = \frac{8+2}{10} = \frac{10}{10} =1$
$\displaystyle c)3x^2-13x+14=0 \\ \\ a=3,~b=-13,~c=14 \\ \\ \Delta=b^2-4ac=(-13)^2-4 \cdot 3 \cdot 14=169-168=1\ \textgreater \ 0 \\ \\ x_1= \frac{-(-13)- \sqrt{1} }{2 \cdot 3} = \frac{13-1}{6} = \frac{12}{6} =2 \\ \\ x_2= \frac{-(-13)+ \sqrt{1} }{2 \cdot 3} = \frac{13+1}{6} = \frac{14}{6} = \frac{7}{3}$
$\displaystyle d)2x^2-9x+10=0 \\ \\ a=2,~b=-9,~c=10 \\ \\ \Delta=b^2-4ac=(-9)^2-4 \cdot 2 \cdot 10=81-80=1\ \textgreater \ 0 \\ \\ x_1= \frac{-(-9)- \sqrt{1} }{2 \cdot 2} = \frac{9-1}{4} = \frac{8}{4} =2 \\ \\ x_2= \frac{-(-9)+ \sqrt{1} }{2 \cdot 2} = \frac{9+1}{4} = \frac{10}{4} = \frac{5}{2}$
$\displaystyle e)5x^2-6x+1=0 \\ \\ a=5,~b=-6,~c=1 \\ \\ \Delta=b^2-4ac=(-6)^2-4 \cdot 5 \cdot1=36-20=16\ \textgreater \ 0 \\ \\ x_1= \frac{-(-6)- \sqrt{16} }{2 \cdot 5} = \frac{6-4}{10} = \frac{2}{10} = \frac{1}{5} \\ \\ x_2= \frac{-(-6)+ \sqrt{16} }{2 \cdot 5} = \frac{6+4}{10} = \frac{10}{10} =1$
$\displaystyle f)4x^2+x-33=0 \\ \\ a=4,~b=1,~c=-33 \\ \\ \Delta=b^2-4ac=1^2-4 \cdot 4 \cdot (-33)=1+528=529\ \textgreater \ 0 \\ \\ x_1= \frac{-1- \sqrt{529} }{2 \cdot 4} = \frac{-1-23}{8} = \frac{-24}{8} =-3 \\ \\x_2= \frac{-1+ \sqrt{529} }{2 \cdot 4} = \frac{-1+23}{8} = \frac{22}{8} = \frac{11}{4}$
$\displaystyle g)x^2-10x-24=0 \\ \\ a=1,~b=-10,~c=-24 \\ \\ \Delta=b^2-4ac=(-10)^2-4 \cdot 1 \cdot (-24)=100+96=196\ \textgreater \ 0 \\ \\ x_1= \frac{-(-10)- \sqrt{196} }{2 \cdot 1} = \frac{10-14}{2} = \frac{-4}{2} =-2 \\ \\ x_2= \frac{-(-10)+ \sqrt{196} }{2 \cdot 1} = \frac{10+14}{2} = \frac{24}{2} =12$
$\displaystyle h)x^2+x-90=0 \\ \\ a=1,~b=1,~c=-90 \\ \\ \Delta=b^2-4ac=1^2-4 \cdot 1 \cdot (-90)=1+360=361\ \textgreater \ 0 \\ \\ x_1= \frac{-1- \sqrt{361} }{2 \cdot 1} = \frac{-1-19}{2} = \frac{-20}{2} =-10 \\ \\ x_2= \frac{-1+ \sqrt{361} }{2 \cdot 1} = \frac{-1+19}{2} = \frac{18}{2} =9$