weldonamy17 weldonamy17

29-11-2020
Mathematics

contestada

f(x)=x^2+12x+35, solve f(x)>0

Respuesta :

mangetsu mangetsu

29-11-2020

Step-by-step explanation:

Solve the eq f(x) =0

x^2+5x+7x+35=0

x*x+5x+7x+5*7=0

x(x+5)+7(x+5)=0

(x+7)*(x+5)=0

x=-7 or x=-5

Now solve f(x) >0

So the function is positive (>0) when

x<-7 and when x>-5

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