Question

Ayudenme a resolver esto x favor que vendria hacer la 2da foto el ejmlpo

Answer 1

${x}^{2} - 2x - 8 = 0 \\ (x - 4)(x + 2) = 0 \\ x - 4 = 0 \: \: \: x + 2 = 0 \\ x_{1} = 4 \: \: \: \: x_{2} = - 2$

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$2 {x}^{2} - 7x - 4 = 0 \\ x = \frac{ - ( - 7)± \sqrt{ {( - 7)}^{2} - 4(2)( - 4)} }{2(2)} \\ x = \frac{7± \sqrt{49 + 32} }{4} \\ x = \frac{7± \sqrt{81} }{4} \\ x = \frac{7±9}{4} \\ x_{1}= \frac{7 + 9}{4} = \frac{16}{4} = 4 \\ x _{2} = \frac{7 - 9 }{4} = \frac{ - 2}{4} = \frac{ - 1}{2}$

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$4{x}^{2} + 11x - 3= 0 \\ x = \frac{ - 11± \sqrt{ {( 11)}^{2} - 4(4)( - 3)} }{2(4)} \\ x = \frac{ - 11± \sqrt{121 + 48} }{8} \\ x = \frac{ - 11± \sqrt{169} }{8} \\ x = \frac{ - 11±13}{8} \\ x_{1}= \frac{ - 11 + 13}{8} = \frac{2}{8} = \frac{1}{4} \\ x _{2} = \frac{ - 11- 13 }{8} = \frac{ - 24}{8} = - 3$

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$6 {x}^{2} - x - 1 = 0 \\ x = \frac{ - ( - 1)± \sqrt{ {( - 1)}^{2} - 4(6)( - 1)} }{2(6)} \\ x = \frac{1± \sqrt{1 - ( - 24)} }{12} \\ x = \frac{1± \sqrt{25} }{12} \\ x = \frac{1±5}{12} \\ x_{1}= \frac{1+ 5}{12} = \frac{6}{12} = \frac{1}{2} \\ x _{2} = \frac{1- 5 }{12} = \frac{ - 4}{12} = \frac{ - 1}{3}$

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$2 {x}^{2} - 11x + 15 = 0 \\ x = \frac{ - ( - 11)± \sqrt{ {( - 11)}^{2} - 4(2)( 15)} }{2(2)} \\ x = \frac{11± \sqrt{121 - 120} }{4} \\ x = \frac{11± \sqrt{1} }{4} \\ x = \frac{11±1}{4} \\ x_{1}= \frac{11 + 1}{4} = \frac{12}{4} = 3 \\ x _{2} = \frac{11 - 1 }{4} = \frac{ 10}{4} = \frac{ 5}{2}$

Espero sea lo que necesitas.