Question

quais são as raízes da equação: x²-14x+48=0 ?

Answer 1

$x^2-14x+48=0$

$\Delta=(-14)^2-4\cdot1\cdot48=196-192=4$

$x=\dfrac{-(-14)\pm\sqrt{4}}{2}=\dfrac{14\pm2}{2}=7\pm1$

$x'=7+1=8$

$x"=7-1=6$

Answer 2

$x^2-14x+48=0$

Por Bháskara:

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

a=1, b=-14, c=48

Δ= b^2 - 4ac
Δ= (-14)^2 - 4*1*48
Δ= 196 -192
Δ= 4

$x = \dfrac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ \\x = \dfrac{-(-14) \pm \sqrt{4}}{2*1} \\ \\ \\ x = \dfrac{14 \pm 2}{2} \\ \\ \\ x' = \dfrac{14 + 2}{2} \\ \\ \\ x' = \dfrac{16}{2} \\ \\ \\ x' = 8 \\ \\ \\ x'' = \dfrac{14 - 2}{2} \\ \\ \\ x'' = \dfrac{12}{2} \\ \\ \\ x'' = 6$

S= {8, 6}