Question

determine as raizes de x2-50x+49=0 ?? mim ajudem pfv

Answer 1

x² - 50x +49 = 0
a = 1
b = -50
c = 49
Δ= b² -4ac
Δ= (-50)² - 4.1.49
Δ= 2500 - 196
Δ= 2304 
x = -b +- √Δ/2a
x = -(-50) +- √2304/2.1
x = 50 +- 48 /2
x' = 98/2 ⇒ x' = 49
x"= 2/2 ⇒x" = 1

S = { 1; 49} 

Answer 2

$x^2-50x+49=0$

Resolvendo por Bháskara.

$x = \dfrac{-b \pm \sqrt{b^2 -4*a*c}}{2*a}$

a=1, b=−50,  c=49

Δ=b2−4ac
Δ=(−50)2−4*(1)*(49)
Δ=2500−196
Δ=2304

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\ x = \dfrac{-(-50) \pm \sqrt{2304}}{2*1} \\ \\ \\ x = \dfrac{50 \pm \sqrt{2304}}{2} \\ \\ \\ x = \dfrac{50 \pm 48}{2} \\ \\ \\ x' = \dfrac{50 + 48}{2} \\ \\ \\ x' = \dfrac{98}{2}\\ \\ \\ x' = 49 \\ \\ \\ \\ x'' = \dfrac{50 - 48}{2} \\ \\ \\ x'' = \dfrac{2}{2} \\ \\ \\ x'' = 1$

S = {49, 1}