Question

Ache o valor de p: (Função quadrática)
400 = 10p² +400p

Answer 1

Dada a expressão:

$\mathsf{400=10p^2+400p}\\\mathsf{10p^2+400p-400=0}\\\mathsf{p^2+40p-40=0}$

Aplicando a formula da função quadrática:

$\boxed{\mathsf{a=1;\:b=40;\:c=-40}}$


$\mathsf{p=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}\\\\\\\mathsf{p=\dfrac{-40\pm \sqrt{40^2-\left[4\cdot \:1\cdot \:\left(-40\right)\right]}}{2\cdot 1}}\\\\\\\mathsf{p=\dfrac{-40\pm \sqrt{1600+160}}{2}}\\\\\\\mathsf{p=\dfrac{-40\pm \sqrt{1760}}{2}}\\\\\\\mathsf{p=\dfrac{-40\pm \sqrt{16\cdot 110}}{2}}\\\\\\\mathsf{p=\dfrac{-40\pm 4\sqrt{110}}{2}}\\\\\\\mathsf{p=-20\pm 2\sqrt{110}}$

$\mathsf{\leadsto p_1=-20+2\sqrt{110}=2\cdot \left(\sqrt{110}-10\right)}\\\\\mathsf{\leadsto p_2=-20-2\sqrt{110}=-2\cdot \left(10+\sqrt{110}\right)}$

Portanto a solução da equação sera:

$\boxed{\mathsf{S=\left\{p\in \mathbb{R}\:,\:p=2\left(\sqrt{110}-10\right)\:\:\:ou\:\:\:p=-2\left(10+\sqrt{110}\right)\right\}}}\: \: \checkmark$