Question

Na equação x2-(p+3)x/4=-2px-7, qual o valor de p para que a soma das raízes seja 11?

Answer 1

$\mathrm{\dfrac{x^2-(p+3)x}{4}=-2px-7\ \to\ x^2-(p+3)x=4(-2px-7)\ \to}\\\\ \mathrm{\to\ x^2-(p+3)x=-8px-28\ \to\ x^2-(p+3)x+8px+28=0\ \to}\\\\ \mathrm{\to\ x^2-px-3x+8px+28=0\ \to\ x^2+(7p-3)x+28=0\ \to}\\\\ \mathrm{\to\ a=1;\ b=7p-3;\ c=28\ \to\ S=-\dfrac{b}{a}\ \to\ 11=-\dfrac{b}{a}\ \to}\\\\ \mathrm{\to\ -11a=b\ \to\ -11.1=7p-3\ \to\ -11+3=7p\ \to}\\\\ \mathrm{\to\ -8=7p\ \to\ \boxed{\mathbf{p=-\dfrac{8}{7}}}}$