Question

Complete o quadrado para calcular x.

2x^2 + 13x +20 = 0
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Quadrado Completo:
(x + )^2 =
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Solução:
x= ou x=

ajudem não consigo resolver

Answer 1

Quadrado da soma de dois termos:

$(a+b)^{2}=a^{2}+2ab+b^{2}$
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$2x^{2}+13x+20=0$

Reescrevendo 2x² na forma a²:

$2x^{2}=(\sqrt{2})^{2}x^{2}=(\sqrt{2}x)^{2}$

Reescrevendo 13x, de modo que fique da forma 2ab:

$13x=2\cdot(\sqrt{2}x)\cdot\dfrac{13}{2\sqrt{2}}$

Com isso, concluímos que $b=\dfrac{13}{2\sqrt{2}}$

Achando b²:

$b^{2}=\left(\dfrac{13}{2\sqrt{2}}\right)^{2}=\dfrac{13^{2}}{\sqrt{8}^{2}}=\dfrac{169}{8}$
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Podemos completar o quadrado somando e subtraindo b² na equação:

$(\sqrt{2}x)^{2}+2\cdot(\sqrt{2}x)\cdot\dfrac{13}{2\sqrt{2}}+\left(\dfrac{13}{2\sqrt{2}}\right)^{2}-\dfrac{169}{8}+20=0$

Finalmente, escrevendo o quadrado da soma:

$\left(x\sqrt{2}+\dfrac{13}{2\sqrt{2}}\right)^{2}+20-\dfrac{169}{8}=0\\\\\\\left(x\sqrt{2}+\dfrac{13}{2\sqrt{2}}\right)^{2}+\dfrac{160}{8}-\dfrac{169}{8}=0\\\\\\\left(x\sqrt{2}+\dfrac{13}{2\sqrt{2}}\right)^{2}-\dfrac{9}{8}=0\\\\\\\left(x\sqrt{2}+\dfrac{13}{2\sqrt{2}}\right)^{2}=\dfrac{9}{8}$

Tirando raiz quadrada dos dois lados da equação:

$\sqrt{\left(x\sqrt{2}+\dfrac{13}{2\sqrt{2}}\right)^{2}}=\sqrt{\dfrac{9}{8}}\\\\\\\left|\left(x\sqrt{2}+\dfrac{13}{2\sqrt{2}}\right)\right|=\dfrac{3}{\sqrt{8}}=\dfrac{3}{2\sqrt{2}}\\\\\\\boxed{\boxed{x\sqrt{2}+\dfrac{13}{2\sqrt{2}}=\pm\dfrac{3}{2\sqrt{2}}}}$

Então, teremos duas possibilidades:

$x\sqrt{2}+\dfrac{13}{2\sqrt{2}}=\dfrac{3}{2\sqrt{2}}\\\\\\x\sqrt{2}=\dfrac{3}{\sqrt{2}}-\dfrac{13}{2\sqrt{2}}\\\\\\x\sqrt{2}=-\dfrac{10}{2\sqrt{2}}\\\\\\x=-\dfrac{5}{\sqrt{2}\cdot\sqrt{2}}\\\\\\\boxed{\boxed{x=-\dfrac{5}{2}}}$

Ou:

$x\sqrt{2}+\dfrac{13}{2\sqrt{2}}=-\dfrac{3}{2\sqrt{2}}\\\\\\x\sqrt{2}=-\dfrac{3}{2\sqrt{2}}-\dfrac{13}{2\sqrt{2}}\\\\\\x\sqrt{2}=-\dfrac{16}{2\sqrt{2}}\\\\\\x=-\dfrac{8}{\sqrt{2}\cdot\sqrt{2}}\\\\\\x=-\dfrac{8}{2}\\\\\\\boxed{\boxed{x=-4}}$