Question

x elevado a 2 +16x= 0 Bhaskara

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Propriedade\,do\,produto\,nulo}\\\sf sejam\,\boldsymbol{a}\,e\,\boldsymbol{b}\,dois\,n\acute umeros\,reais\\\sf quaisquer.\\\sf se\,\boldsymbol{a\cdot b=0}\,ent\tilde ao~\boldsymbol{a=0}\,ou\,\boldsymbol{b=0}\\\underline{\rm Form\acute ula\,resolutiva\,para\,equac_{\!\!,}\tilde oes}\\\underline{\rm do\,2^o~grau\,(f\acute ormula\,de\,Bh\acute askara)}\\\sf seja\,ax^2+bx+c=0\,com\,a,b\,e\,c\in\mathbb{R}\,e\,a\ne0.\\\sf a\,soluc_{\!\!,}\tilde ao\,desta\,equac_{\!\!,}\tilde ao\,\acute e\,dada\,por\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a},onde,\Delta=b^2-4ac\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Resoluc_{\!\!,}\tilde ao\,pela\,propriedade}\\\underline{\rm\,do\,produto\,nulo:}\\\sf x^2+16x=0\\\sf x\cdot(x+16)=0\\\sf x=0\\\sf x+16=0\\\sf x=-16\\\sf S=\{0,-16\}\\\underline{\rm Resoluc_{\!\!,}\tilde ao\,pela\,f\acute ormula\,de\,Bh\acute askara:}\\\sf x^2+16x=0\implies \begin{cases}\sf a=1\\\sf b=16\\\sf c=0\end{cases}\\\sf \Delta=b^2-4ac\\\sf\Delta=16^2-4\cdot1\cdot0\\\sf\Delta=256\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(16)\pm\sqrt{256}}{2\cdot1}\\\\\sf x=\dfrac{-16\pm16}{2}\begin{cases}\sf x_1=\dfrac{-16+16}{2}=\dfrac{0}{2}=0\\\\\sf x_2=\dfrac{-16-16}{2}=-\dfrac{32}{2}=-16\end{cases}\\\sf S=\{0,-16\}\end{array}}$