Question

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a) 5x² - 30 = x²+34
b) (x - 3) (x+3) + 25= 0
c) ( x - 2)² + 6x = 4
d) x² - 9x + 8 = 0
e) x² + 7x + 10 = 0
f) 6x² - 5x + 1 = 0

Answer 1

$\displaystyle a).5x^2-30=x^2+34 \\ 5x^2-x^2=34+30 \\ 4x^2=64 \\ x^2= \frac{64}{4} \\ x^2=16 \\ x= \sqrt{16} \\ x=4;x=-4$

$b).(x-3)(x+3)+25=0 \\ x^2-9+25=0 \\ x^2+16=0 \\ x^2=0-16 \\ x^2=-16 \\ x= \sqrt{-16} \\ x=4;x=-4$

$c).(x-2)^2+6x=4 \\ x^2-2 \cdot x \cdot 2+2^2+6x=4 \\ x^2-4x+4+6x=4 \\ x^2+2x+4=4 \\ x^2+2x+4-4=0 \\ x^2+2x=0 \\ x(x+2)=0 \\ x=0;x=-2$

$\displaystyle d).x^2-9x+8=0 \\ a=1,b=-9,c=8 \\ D=b^2-4ac=9^2-4 \cdot 1 \cdot 8=81-32=49\ \textgreater \ 0 \\ x_1= \frac{9+ \sqrt{49} }{2 \cdot 1} = \frac{9+7}{2} = \frac{16}{2} =8 \\ \\ x_2= \frac{9- \sqrt{49} }{2 \cdot 1} = \frac{9-7}{2} = \frac{2}{2} =1$

$\displaystyle e).x^2+7x+10=0 \\ a=1,b=7,c=10 \\ D=b^2-4ac=7^2-4 \cdot 1 \cdot 10=49-40=9\ \textgreater \ 0 \\ x_1= \frac{-7+ \sqrt{9} }{2 \cdot 1} = \frac{-7+3}{2} = \frac{-4}{2} =-2 \\ \\ x_2= \frac{-7- \sqrt{9} }{2 \cdot 1} = \frac{-7-3}{2} = \frac{-10}{2} =-5$

$\displaystyle f).6x^2-5x+1=0 \\ a=6,b=-5,c=1 \\ D=b^2-4ac=5^2-4 \cdot 6 \cdot 1=25-24=1\ \textgreater \ 0 \\ x_1= \frac{5+ \sqrt{1} }{2 \cdot 6} = \frac{5+1}{12} = \frac{6}{12} = \frac{1}{2} \\ \\ x_2= \frac{5- \sqrt{1} }{2 \cdot 6} = \frac{5-1}{12} = \frac{4}{12} = \frac{1}{3}$