Find the roots of quad eq. 1\3x²-√11x+1=0 if they exist using quad formula About the author Margaret
Answer: Step-by-step explanation: [tex]\frac{1}{3}[/tex] x² – √11 x + 1 = 0 D = b² – 4ac = (- √11)² – 4( [tex]\frac{1}{3}[/tex] ) (1) = 11 – 1[tex]\frac{1}{3}[/tex] = 9[tex]\frac{3}{4}[/tex] = [tex]\frac{39}{4}[/tex] = ( [tex]\frac{\sqrt{39} }{2}[/tex] )² [tex]x_{12}[/tex] = [tex]\frac{1}{2a}[/tex] ( – b ± √D ) [tex]x_{1}[/tex] = [tex]\frac{3}{2}[/tex] ( √11 – [tex]\frac{\sqrt{39} }{2}[/tex] ) = [tex]\frac{3\sqrt{11} }{2}[/tex] – [tex]\frac{3\sqrt{39} }{4}[/tex] = [tex]\frac{3}{4}[/tex] ( 2√11 – √39 ) [tex]x_{2}[/tex] = [tex]\frac{3}{4}[/tex] ( 2√11 + √39 ) Reply
Answer:
Step-by-step explanation:
[tex]\frac{1}{3}[/tex] x² – √11 x + 1 = 0
D = b² – 4ac = (- √11)² – 4( [tex]\frac{1}{3}[/tex] ) (1) = 11 – 1[tex]\frac{1}{3}[/tex] = 9[tex]\frac{3}{4}[/tex] = [tex]\frac{39}{4}[/tex] = ( [tex]\frac{\sqrt{39} }{2}[/tex] )²
[tex]x_{12}[/tex] = [tex]\frac{1}{2a}[/tex] ( – b ± √D )
[tex]x_{1}[/tex] = [tex]\frac{3}{2}[/tex] ( √11 – [tex]\frac{\sqrt{39} }{2}[/tex] ) = [tex]\frac{3\sqrt{11} }{2}[/tex] – [tex]\frac{3\sqrt{39} }{4}[/tex] = [tex]\frac{3}{4}[/tex] ( 2√11 – √39 )
[tex]x_{2}[/tex] = [tex]\frac{3}{4}[/tex] ( 2√11 + √39 )