Question

b) 16x-3(x+1)=5x
c)2,5(x-4)-6x=3-x
d) x² -3x-10=0
e)2x² +7x-8=0
f) -x² +10x+2=0
Matematica cl.9 pg.9 ex. 8

Answer 1

$\displaystyle a).16x-3(x+1)=5x \\ 16x-3x-3=5x \\ 16x-3x-5x=3 \\ 8x=3 \\ x= \frac{3}{8} \\ \\ b).2,5(x-4)-6x=3-x \\ 2,5x-10-6x=3-x \\ 2,5x-6x+x=3+10 \\ -2,5x=13 \\ x=- \frac{13}{2,5} \\ x=-5,2$

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

$\displaystyle d).2x^2+7x-8=0 \\ a=2,b=7,c=-8 \\ \Delta=b^2-4ac=7^2-4 \cdot 2 \cdot (-8)=49+64=113\ \textgreater \ 0 \\ x_1= \frac{-7+ \sqrt{113} }{2 \cdot 2} = \frac{-7+ \sqrt{113} }{4} \\ \\ x_2= \frac{-7- \sqrt{113} }{2 \cdot 2} = \frac{-7- \sqrt{113} }{4}$

$\displaystyle e).-x^2+10x+2=0 \\ a=-1,b=10,c=2 \\ \Delta=b^2-4ac=10^2-4 \cdot (-1) \cdot 2=100+8=108 \ \textgreater \ 0 \\ x_1= \frac{-10+ \sqrt{108} }{2 \cdot (-1)} = \frac{-10+6 \sqrt{3} }{-2} = \frac{-2(5-3 \sqrt{3} )}{-2} =5-3 \sqrt{3} \\ \\ x_2= \frac{-10- \sqrt{108} }{2 \cdot (-1)} = \frac{-10-6 \sqrt{3} }{-2} = \frac{-2(5+3 \sqrt{3} )}{-2} =5+3 \sqrt{3}$