Question

$64y {}^{2} \: + \: 48y \: + 9 =$
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Answer 1

Resposta:

S={-3/8}, raiz duplas

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~64y^{2}+48y+9=0~~\\e~comparando~com~(a)y^{2}+(b)y+(c)=0,~temos~a=64{;}~b=48~e~c=9\\\\\Delta=(b)^{2}-4(a)(c)=(48)^{2}-4(64)(9)=2304-(2304)=0\\\\y^{'}=y^{''}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(48)-\sqrt{0}}{2(64)}=\frac{-48-0}{128}=\frac{-48\div16}{128\div16}=-\frac{3}{8}$