Question

Rezolvati in R ecuatile si determinatile ce tip de numere ,rationale sau irationale ,sint solutile ei:
a)x²-3x-10=0
b)2x²+7x-8=0
c)-x²+10x+2=0

Answer 1

$\displaystyle a).x^2-3x-10=0 \\ a=1,b=-3,c=-10 \\ \Delta=b^2-4ac=3^2-4 \cdot 1 \cdot (-10)=9+40=49\ \textgreater \ 0 \\ x_1= \frac{3+ \sqrt{49} }{2 \cdot1} = \frac{3+7}{2} = \frac{10}{2} =5 \\ \\ x_2= \frac{3- \sqrt{49} }{2 \cdot 1} = \frac{3-7}{2} = \frac{-4}{2} =-2$

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

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