Question

Determine as raízes de cada equação do 2° grau.

X2 - 3x -10 = 0

X2 + 3X - 4 = 0​

Answer 1

1- x² + 3x - 10 = 0

Delta = 3² - 4.(1).(-10)

Delta = 9 + 40

Delta = 49

x = -3 ± √49/2.1

x = -3 ± 7/2

x¹ = -3 + 7/2

x¹ = 2

x² = -3 - 7/2

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2- x² + 3x - 4 = 0

Delta = 3² - 4.(1).(-4)

Delta = 9 + 16

Delta = 25

x = -3 ± √25/2.1

x = -3 ± 5/2

x¹ = -3 + 5/2

x¹ = 1

x² = -3 - 5/2

x² = 4

Espero ter ajudado!

Bons estudos :)

Answer 2

Resposta:

X2 - 3x -10 = 0

a = 1

b = -3

c = -10

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-3) +- \sqrt{(-3)^{2} - 4 * 1 * -10} }{2*1}\\\\x = \frac{3+-\sqrt{9+40} }{2}\\\\x = \frac{3+-\sqrt{49} }{2}\\\\x = \frac{3+-7}{2}\\\\x' = \frac{3+7}{2}=\frac{10}{2}=5 \\\\x'' = \frac{3-7}{2}=\frac{-4}{2} =-2$

X2 + 3X - 4 = 0​

a = 1

b = 3

c = -4

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(3) +- \sqrt{(3)^{2} - 4 * 1 * -4} }{2*1}\\\\x = \frac{-3+-\sqrt{9+16} }{2}\\\\x = \frac{-3+-\sqrt{25} }{2}\\\\x = \frac{-3+-5}{2}\\\\x' = \frac{-3+5}{2}=\frac{2}{2}=1\\\\x'' = \frac{-3-5}{2}=\frac{-8}{2}=-4$