Question

exerctul 3 si 4 va rog! e urgent

Answer 1

[tex]1) x^{2} -7x+6=0\\ a=1 \\b=-7\\c=6\\ \Delta=(-7)^{2}-4*1*6=49-24=25\\ \sqrt{\Delta} = \sqrt{25}=5 \\x_{1}= \frac{-b- \sqrt{\Delta} }{2a}= \frac{7-5}{2}= \frac{2}{2}=1\\ x_{2}= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{7+5}{2}= \frac{12}{2}=6\\ S=[1, 6] \\ 2) 4 x^{2} -12x+5=0\\ a=4\\b=-12\\c=5\\ \Delta=(-12)^2-4*4*5=144-80=64\\ x_{1}= \frac{-b- \sqrt{\Delta} }{2a}= \frac{12-8}{2*4}= \frac{4}{8}= \frac{1}{2}\\ x_{2}= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{12+8}{2*4}=\frac{20}{8}=\frac{5}{2}\\ [/tex] [tex]S=[ \frac{1}{2} ; \frac{5}{2} ]\\ 3) x^{2} -3 \sqrt{3}x+6=0\\ a=1\\b=-3 \sqrt{3}\\c=6\\ \Delta=b^2-4ac=(-3\sqrt{3})^2-4*1*6=27-24=3\\ x_{1}= \frac{-b- \sqrt{\Delta} }{2a}= \frac{3 \sqrt{3}- \sqrt{3}}{2}= \frac{2 \sqrt{3} }{2}= \sqrt{3}\\ x_{2}= \frac{-b+ \sqrt{\Delta} }{2a}=\frac{3 \sqrt{3}+ \sqrt{3}}{2}= \frac{4 \sqrt{3} }{2}=2 \sqrt{3}\\ S=[ \sqrt{3}; 2 \sqrt{3} ] \\[/tex]
[tex]4) x^{2} -4 \sqrt{2}x-6=0\\ a=1\\b=-4\sqrt{2}\\c=-6\\ \Delta=(-4\sqrt{2})^{2}-4*1*(-6)=32+24=56\\ x_{1}= \frac{-b- \sqrt{\Delta} }{2a}= \frac{4\sqrt{2}-\sqrt{56}}{2}= \frac{4\sqrt{2}-\sqrt{4*14}}{2}= \\=\frac{4\sqrt{2}-2 \sqrt{14} }{2}= \frac{2(2\sqrt{2}-\sqrt{14})}{2}= 2\sqrt{2}-\sqrt{14} \\ x_{2}= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{4\sqrt{2}+\sqrt{56}}{2}= \frac{4\sqrt{2}+\sqrt{4*14}}{2}= \\=\frac{4\sqrt{2}+2 \sqrt{14} }{2}= \frac{2(2\sqrt{2}+\sqrt{14})}{2}= 2\sqrt{2}+\sqrt{14} \\ [/tex]
$S=[2\sqrt{2}-\sqrt{14}; 2\sqrt{2}+\sqrt{14}]$