• Matéria: Matemática
  • Autor: paulajoselia
  • Perguntado 9 anos atrás

x²-7x+10=0

x²-6x+9=o

formula de báskara

Respostas

respondido por: GabrielAw9
0
Δ = b² -4ac
Δ = (-7)² -4 *1 * 10
Δ = 49 -40
Δ = 9
x' = (-b + √∆)/ 2a
(-(-7) + 3)/2 * 1
(7 + 3)/2
10/2
5
x'' = (-b - √∆)/ 2a
(-(-7) - 3)/2 * 1
(7 - 3)/2
4/2
2




Δ = b² -4ac
Δ = (-6)² -4 *1 * 9
Δ = 36 -36
Δ = 0
x' = (-b + √∆)/ 2a
(-(-6) + 0)/2 * 1
(6 + 0)/2
6/2
3
x'' = (-b - √∆)/ 2a
(-(-6) - 0)/2 * 1
(6 - 0)/2
6/2
3


respondido por: FranciscoRamon
1
 \frac{-b  \frac{+}{} \sqrt{(b)²-4.a.c} }{2.a}


a) \frac{-(-7)  \frac{+}{} \sqrt{(-7)^{2}-4.1.10} }{2.1}
 \frac{7  \frac{+}{} \sqrt{49-40} }{2}


 \frac{7\frac{+}{} \sqrt{9} }{2}


 \frac{7\frac{+}{} 3 }{2}

x' = (7+ 3)/2 => x' = 10/2 = 5
x'' = (7 -3)/2 => x' = 4/2 = 2

b) \frac{-(-6)  \frac{+}{} \sqrt{(-6)^{2}-4.1.9} }{2.1}


 \frac{6\frac{+}{} \sqrt{36-36} }{2}


 \frac{6\frac{+}{} \sqrt{0} }{2}


 \frac{6\frac{+}{} 0 }{2}

x = 6/2 => x = 3




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