Question

calcule o valor das raízes x' e x'' da função quadrática f(x)1x²-9x+8

Answer 1

f(x)=x²-9x+8
x²-9x+8=0
a= 1 b= -9 c-8
Δ= b²-4ac
Δ= (-9)²-4·1·8
Δ=81-32
Δ=49

x= -b+/-√Δ/2a
x= - (-9) +/-√49/2·1
x= 9+/-7/2

x'= 8
x''= 1

S:[1,8]

Answer 2

f(x)= 1x²-9x+8
$\boxed{A = 1 , B = -9 , C = 8}$

Δ = b² -4ac
Δ = (-9) ² -4.1.8
Δ = 81 - 32
Δ = 49

x = -b  ± √Δ
          2a

x = -(-9)  ± √49
            2.1

x = 9  ± 7
         2

$\boxed{x' =\frac{9+7}{2} = \frac{16}{2} = 8}$

$\boxed{x''= \frac{9-7}{2} = \frac{2}{2} = 1}$