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How do you solve these algebra questains?
2m^2-7m-3=0
4b^2+8b+7=4
5√80a^2
-6√150r
2/√3
√6/5√3

Sagot :

[tex]2m^2-7m-3=0\\a=2;\ b=-7;\ c=-3\\\Delta=b^2-4ac\to\Delta=(-7)^2-4\cdot2\cdot(-3)=49+24=73\\\\x=\frac{-b\pm\sqrt\Delta}{2a}\to \boxed{x=\frac{7\pm\sqrt{73}}{2\cdot2}=\frac{7\pm\sqrt{73}}{4}}[/tex]


[tex]4b^2+8b+7=4\\4b^2+8b+7-4=0\\4b^2+8b+3=0\\4b^2+2b+6b+3=0\\2b(2b+1)+3(2b+1)=0\\(2b+1)(2b+3)=0\iff2b+1=0\ or\ 2b+3=0\\2b=-1\ or\ 2b=-3\\\boxed{b=-\frac{1}{2}\ or\ b=-\frac{3}{2}}[/tex]


[tex]5\sqrt{80a^2}=5\sqrt{80}\cdot\sqrt{a^2}=5\sqrt{16\cdot5}\cdot|a|=5\cdot\sqrt{16}\cdot\sqrt5\cdot|a|\\\\=5\cdot4\cdot\sqrt5\cdot|a|=\boxed{20|a|\sqrt5}[/tex]


[tex]-6\sqrt{150r}=-6\sqrt{25\cdot6r}=-6\sqrt{25}\cdot\sqrt{6r}=-6\cdot5\sqrt{6r}=\boxed{-30\sqrt{6r}}[/tex]


[tex]\frac{2}{\sqrt3}=\frac{2}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\boxed{\frac{2\sqrt3}{3}}[/tex]


[tex]\frac{\sqrt6}{5\sqrt3}=\frac{1}{5}\cdot\frac{\sqrt6}{\sqrt3}=\frac{1}{5}\sqrt\frac{6}{3}=\frac{1}{5}\sqrt2=\boxed{\frac{\sqrt2}{5}}[/tex]
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