Question

dada a função f:R definida por f(x)=x2-5x+6 calcule os valore reais de X para que se tenha f(x)=12

Answer 1

$x^{2} -5x+6=12 \\ x^{2} -5x-6=0 \\ x= \frac{-b+- \sqrt{ b^{2}-4.a.c} }{2.a} \\ \\ x= \frac{5+- \sqrt{ -5^{2}-4.1.-6} }{2.1} \\ \\ x= \frac{5+- \sqrt{49} }{2.1} \\ \\ x= \frac{5+7 }{2} \\ \\ x=6 \\ \\ x= \frac{5-7 }{2.1} \\ \\ x= \frac{5-7 }{2.1} \\ \\ x=-1$

x=6 e x=-1

Answer 2

f(x)=x^2-5x+6
x^2-5x+6=12
x^2-5x+6-12
x^2-5x-6=0
delta=b^2-4ac
delta=(-5)^2-4.1.(-6)
delta=25+24
delta=49
x=-b-/+ raiz de delta/2a
x=-(-5)-7/2.1
x=5-7/2
x=-2/2
x=-1
x=-(-5)+7/2.1
x=5+7/2
x=12/2
x=6