Question

Determine k ∈ a R para que kx² - 4x + k – 3 < 0, para todo x real.
Se souber faça. Caso contrário, nem tente.

Answer 1

$\boxed{\begin{array}{l}\sf kx^2-4x+k-3<0\Longleftrightarrow k<0~e~\Delta<0.\\\sf \Delta=b^2-4ac\\\sf\Delta=(-4)^2-4\cdot k\cdot(k-3)\\\sf\Delta=16-4k^2+12k\end{array}}$

$\boxed{\begin{array}{l}\sf\Delta<0\implies -4k^2+12k+16<0\\\underline{\boldsymbol{Fac_{\!\!,}a}}\\\sf f(k)=-4k^2+12k+16\\\underline{\rm zeros~da~func_{\!\!,}\tilde ao~f(k):}\\\sf -4k^2+12k+16=0\\\sf\Delta=b^2-4ac\\\sf\Delta=12^2-4\cdot(-4)\cdot16\\\sf\Delta=144+256\\\sf\Delta=400\\\sf k=\dfrac{-b\pm\sqrt{\Delta}}{2a}\end{array}}$

$\boxed{\begin{array}{l}\sf k=\dfrac{-12\pm\sqrt{400}}{2\cdot(-4)}\\\\\sf k=\dfrac{-12\pm20}{-8}\begin{cases}\sf k_1=\dfrac{-12+20}{-8}=\dfrac{8}{-8}=-1\\\\\sf k_2=\dfrac{-12-20}{-8}=\dfrac{-32}{-8}=4\end{cases}\\\\\sf f(k)<0\Longleftrightarrow k<-1~ou~k>4\end{array}}$

$\large\boxed{\begin{array}{l}\sf fazendo~a~intersecc_{\!\!,}\tilde ao~das\\\sf~duas~condic_{\!\!,}\tilde oes~temos:\\\sf kx^2-4x+k-3<0\Longleftrightarrow k<-1\end{array}}$