Question

F)x²-12x-85=0 S=
Me ajudemmmm​

Answer 1

A) S = {(0,15)}
B) S = {(9,-9)}
C) S = {(-5,1)}
D) S = {(1/2,4)}
E) S = {(-4)}
F) S = {(17,-5)}

Recomendo fazer pelo método de fatoração.

Answer 2

a) x² - 15x = 0

a = 1        b = -15        c = 0

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-15)\pm\sqrt{(-15)^2-4\cdot1\cdot0}}{2\cdot1}=\frac{15\pm\sqrt{225-0}}{2}=\frac{15\pm\sqrt{225}}{2}=\frac{15\pm15}{2}\\\\x'=\frac{15+15}{2}=\frac{30}{2}=15\\\\x''=\frac{15-15}{2}=\frac{0}{2}=0\\\\S=\{0,\ 15\}$

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b) x² - 81 = 0

a = 1        b = 0        c = -81

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-0\pm\sqrt{0^2-4\cdot1\cdot(-81)}}{2\cdot1}=\frac{0\pm\sqrt{0+324}}{2}=\frac{0\pm\sqrt{324}}{2}=\frac{0\pm18}{2}\\\\x'=\frac{0+18}{2}=\frac{18}{2}=9\\\\x''=\frac{0-18}{2}=\frac{-18}{2}=-9\\\\S=\{\pm9\}$

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c) x² + 4x - 5 = 0

a = 1        b = 4       c = -5

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot(-5)}}{2\cdot1}=\frac{-4\pm\sqrt{16+20}}{2}=\frac{-4\pm\sqrt{36}}{2}=\frac{-4\pm6}{2}\\\\x'=\frac{-4+6}{2}=\frac{2}{2}=1\\\\x''=\frac{-4-6}{2}=\frac{-10}{2}=-5\\\\S=\{-5, \ 1\}$

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d) 2x² - 9x + 4 = 0

a = 2        b = -9        c = 4

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-9)\pm\sqrt{(-9)^2-4\cdot2\cdot4}}{2\cdot2}=\frac{9\pm\sqrt{81-32}}{4}=\frac{9\pm\sqrt{49}}{4}=\frac{9\pm7}{4}\\\\x'=\frac{9+7}{4}=\frac{16}{4}=4\\\\x''=\frac{9-7}{4}=\frac{2^{:2}}{4^{:2}}=\frac{1}{2}\\\\S=\Big\{\frac{1}{2}, \ 4\Big\}$

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e) x² + 8x + 16 = 0

a = 1        b = 8        c = 16

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-8\pm\sqrt{8^2-4\cdot1\cdot16}}{2\cdot1}=\frac{-8\pm\sqrt{64-64}}{2}=\frac{-8\pm\sqrt{0}}{2}=\frac{-8\pm0}{2}\\\\x'=\frac{-8+0}{2}=\frac{-8}{2}=-4\\\\x''=\frac{-8-0}{2}=\frac{-8}{2}=-4\\\\S=\{-4\}$

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f) x² - 12x - 85 = 0

a = 1        b = -12        c = -85

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-12)\pm\sqrt{(-12)^2-4\cdot1\cdot(-85)}}{2\cdot1}=\frac{12\pm\sqrt{144+340}}{2}=\frac{12\pm\sqrt{484}}{2}=\frac{12\pm22}{2}\\\\x'=\frac{12+22}{2}=\frac{34}{2}=17\\\\x''=\frac{12-22}{2}=\frac{-10}{2}=-5\\\\S=\{-5,\ 17\}$