Question

-x²+14x-49=0
Me ajudem nessa por favor!!

Answer 1

$-x^2+14x-49=0 ~~~~\times(-1)\\ \\ x^2-14x+49=0\\ \\ a=1\\b=-14 \\ c=49\\ \\ \Delta=b^2-4ac\\ \Delta=(-14)^2-4(1)(49)\\ \Delta=196-196\\ \Delta=0\\ \\ x-{-b\pm\sqrt{\Delta} \over2a}={-(-14)\pm\sqrt{0} \over2(1)}={14\pm0\over2}\\ \\ x'=x"={14\over2}=7\\ \\ S=\{7\}$

Answer 2

Olá

- x² + 14x - 49 = 0

a = - 1

b = 14

c = - 49

1) Calculando o Δ:

Δ = b2 - 4.a.c

Δ = 142 - 4 . -1 . -49

Δ = 196 - 4. -1 . -49

Δ = 0

Há 1 raiz real.

2) Aplicando Bhaskara:

Neste caso, x' = x'':

x = (-b +- √Δ)/2a

x' = (-14 + √0)/2.-1

x' = -14 / -2

x' = 7

x'' = (-14 - √0)/2.-1

x'' = -14 / -2

x'' = 7

S={x'=7 e x"=7}

Espero ter ajudado