Question

resolva as equações e indique se as raízes são iguais ou diferentes

a)x2+x-12=0

b)x2-14x+49=0

Answer 1

a) x² + x - 12 = 0

Δ = 1 -4.1.-12
Δ = 1 + 48
Δ = 49

$x = \frac{-b +- \sqrt{delta} }{2.1}$

$x' = \frac{-1 + 7}{2}$
$x' = 3$

$x" = \frac{-1-7}{2}$
$x'' = -4$

b) x² - 14x + 49 = 0

Δ = 196 - 4.1.49
Δ = 196 + 196
Δ = 0

$x = \frac{14+- \sqrt{delta} }{2}$
$x = \frac{14 +- 0}{2}$
$x' = x'' = 7$

a) diferentes 
b) iguais

Answer 2

a) $x^{2} +x-12=0$

Δ= $1+48$
Δ= 49

$x'= \frac{-1+ \sqrt{49} }{2} \\ x'= \frac{-1+7}{2} \\ x'= \frac{6}{2} \\ x'=3$

$x''= \frac{-1- \sqrt{49} }{2} \\ x''= \frac{-1-7}{2} \\ x''= \frac{-8}{2} \\ x''=-4$

Portanto, raízes diferentes.

b) $x^{2} -14x+49=0$

Δ=196-196
Δ=0

$x'= \frac{14+ \sqrt{0} }{2} = \frac{14}{2}=7$

$x''= \frac{14+ \sqrt{0} }{2} = \frac{14}{2}=7$

Portanto, as raízes são iguais.