Matematică, întrebare adresată de MaximLesnik, 8 ani în urmă

x²-14x+13=0 x²+12x+35=0 7x²-2x-14=0 Dau coroana...

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Răspuns de QuaTeam
5

x²-14x+13=0

\frac{-\left(-14\right)+\sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:13}}{2\cdot \:1}

=\frac{14+\sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:13}}{2\cdot \:1}

\sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:13}

\left(-14\right)^2

4\cdot \:1\cdot \:13=52

=\sqrt{196-52}

=\sqrt{144}

=14+\sqrt{144}

=\frac{14+\sqrt{144}}{2\cdot \:1}

=\frac{14+\sqrt{144}}{2}

\sqrt{144}=12

=\frac{14+12}{2}

=\frac{26}{2}

=13

\frac{-\left(-14\right)-\sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:13}}{2\cdot \:1}

14-\sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:13}

\sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:13}

\left(-14\right)^2=196

4\cdot \:1\cdot \:13=52

=\sqrt{196-52}

=\sqrt{144}

=14-\sqrt{144}

=\frac{14-\sqrt{144}}{2\cdot \:1}

\sqrt{144}=12

=\frac{14-12}{2}

=\frac{2}{2}

=1

x=13,\:x=1

b) x²+12x+35=0

\frac{-12+\sqrt{12^2-4\cdot \:1\cdot \:35}}{2\cdot \:1}

\sqrt{12^2-4\cdot \:1\cdot \:35}

=\sqrt{144-4\cdot \:1\cdot \:35}

=\sqrt{144-140}

=\sqrt{4}

=\sqrt{2^2}

=2\\

=\frac{-12+2}{2}

=\frac{-10}{2}

=-5

\frac{-12-\sqrt{12^2-4\cdot \:1\cdot \:35}}{2\cdot \:1}

\sqrt{12^2-4\cdot \:1\cdot \:35}

=\sqrt{144-4\cdot \:1\cdot \:35}

=\sqrt{144-140}

=\sqrt{4}

=\frac{-12-\sqrt{4}}{2\cdot \:1}

=\frac{-12-\sqrt{4}}{2}

\sqrt{4}=2

=\frac{-12-2}{2}

=\frac{-14}{2}

=-\frac{14}{2}

=-7

x=-5,\:x=-7

c) 7x²-2x-14=0

\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:7\left(-14\right)}}{2\cdot \:7}

=\frac{2+\sqrt{\left(-2\right)^2+4\cdot \:7\cdot \:14}}{2\cdot \:7}

2+\sqrt{\left(-2\right)^2+4\cdot \:7\cdot \:14}

\sqrt{\left(-2\right)^2+4\cdot \:7\cdot \:14}

\left(-2\right)^2=4

4\cdot \:7\cdot \:14=392

=\sqrt{4+392}

=\sqrt{396}

=2+\sqrt{396}

=\frac{2+\sqrt{396}}{2\cdot \:7}

=\frac{2+\sqrt{396}}{14}

\sqrt{396}

2^2\cdot \:3^2\cdot \:11

=\sqrt{2^2\cdot \:3^2\cdot \:11}

=\sqrt{11}\sqrt{2^2}\sqrt{3^2}

=2\cdot \:3\sqrt{11}

=6\sqrt{11}

=\frac{2+6\sqrt{11}}{14}

2+6\sqrt{11}

=2\cdot \:1+2\cdot \:3\sqrt{11}

=2\left(1+3\sqrt{11}\right)

=\frac{1+3\sqrt{11}}{7}

\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:7\left(-14\right)}}{2\cdot \:7}

2-\sqrt{\left(-2\right)^2+4\cdot \:7\cdot \:14}

\left(-2\right)^2=4

4\cdot \:7\cdot \:14=392

=\sqrt{4+392}

=\sqrt{396}

=2-\sqrt{396}

=\frac{2-\sqrt{396}}{2\cdot \:7}

=\frac{2-\sqrt{396}}{14}

\frac{2\left(1-3\sqrt{11}\right)}{14}

=\frac{1-3\sqrt{11}}{7}

x=\frac{1+3\sqrt{11}}{7},\:x=\frac{1-3\sqrt{11}}{7}

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