Question

Resolva as equações a seguir :

a)9x²=36

b)x²+28=2x²+3

c)15x²+5=31x²+4

d)2x²+2/6+6x²-2/12=7/2

e)x²-1/12=4

f)-3x²/6-4=-36

g)13x²-5=6-2(2x²-3)

Answer 1

Temos equações do 2º grau incompletas na forma ax² + c = 0. Os coeficientes são:

• "a" que multiplica x²
• "b" que multiplica x
• "c" que é o termo independente

Para encontrar os valores de x neste caso de equação incompleta, a ideia é isolar a incógnita, extrair a raiz quadrada dos membros e assim obter dois valores. Acompanhe as resoluções:

$~~$

A )

$\begin{array}{l}\sf 9x^2=36\\\\ \sf x^2=\dfrac{36}{9}\\\\ \sf x^2=4\\\\ \sf \sqrt{x^2}=\pm~\sqrt{4}\\\\ \sf x=\pm~2\\\\ \sf \therefore~~x'=2\quad ou\quad x''=-2\end{array}$

Assim o conjunto solução é:

$\large\boxed{\begin{array}{l}\sf S=\Big\{-2~~;~~2\Big\}\end{array}}$

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B )

$\begin{array}{l}\sf x^2+28=2x^2+3\\\\ \sf 28-3=2x^2-x^2\\\\ \sf 25=x^2\\\\ \sf \sqrt{x^2}=\pm~\sqrt{25}\\\\ \sf x=\pm~5\\\\ \sf \therefore~~x'=5\quad ou\quad x''=-5\end{array}$

Assim o conjunto solução é:

$\large\boxed{\begin{array}{l}\sf S=\Big\{-5~~;~~5\Big\}\end{array}}$

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C )

$\begin{array}{l}\sf 15x^2+5=31x^2+4\\\\ \sf 5-4=31x^2-15x^2\\\\ \sf 1=16x^2\\\\ \sf x^2=\dfrac{1}{16}\\\\ \sf \sqrt{x^2}=\pm~\sqrt{\dfrac{1}{16}}\\\\ \sf x=\pm~\dfrac{\sqrt{1}}{\sqrt{16}} \\\\ \sf x=\pm~\dfrac{1}{4}\\\\ \sf \therefore~~x'=\dfrac{1}{4}\quad ou\quad x''=-\dfrac{1}{4}\end{array}$

Assim o conjunto solução é:

$\large\boxed{\begin{array}{l}\sf S=\bigg\{-\dfrac{1}{4}~~;~~\dfrac{1}{4}\bigg\}\end{array}}$

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D )

$\begin{array}{l}\sf \dfrac{2x^2+2}{6}+\dfrac{6x^2-2}{12}=\dfrac{7}{2}\\\\ \Rightarrow~~\sf MMC(6,12,2)=12\\\\ \sf 12\div 6\cdot(2x^2+2)+12\div 12\cdot(6x^2-2)=12\div 2\cdot (7)\\\\ \sf 4x^2+4+6x^2-2=42\\\\ \sf 4x^2+6x^2=42-4+2\\\\ \sf 10x^2=40\\\\ \sf x^2=\dfrac{40}{10}\\\\ \sf x^2=4\\\\ \sf \sqrt{x^2}=\pm~\sqrt{4}\\\\ \sf x=\pm~2\\\\ \sf \therefore~~x'=2\quad ou\quad x''=-2\end{array}$

Assim o conjunto solução é:

$\large\boxed{\begin{array}{l}\sf S=\Big\{-2~~;~~2\Big\}\end{array}}$

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E )

$\begin{array}{l}\sf \dfrac{x^2-1}{12}=4\\\\ \sf x^2-1=12\cdot4\\\\\sf x^2-1=48\\\\ \sf x^2=48+1\\\\ \sf x^2=49\\\\ \sf \sqrt{x^2}=\pm~\sqrt{49}\\\\ \sf x=\pm~7\\\\ \sf \therefore~~x'=7\quad ou\quad x''=-7\end{array}$

Assim o conjunto solução é:

$\large\boxed{\begin{array}{l}\sf S=\Big\{-7~~;~~7\Big\}\end{array}}$

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F )

$\begin{array}{l}\sf -\dfrac{3x^2}{6}-4=-36\\\\ \sf -\dfrac{x^2}{2}-4=-36\\\\ \sf -x^2+2\cdot(-4)=2\cdot(-36)\\\\ \sf -x^2-8=-72\\\\ \sf -x^2=-72+8\\\\ \sf -x^2=-64~~\cdot(-1)\\\\ \sf x^2=64\\\\ \sf \sqrt{x^2}=\pm~\sqrt{64}\\\\ \sf x=\pm~8\\\\ \sf \therefore~~x'=8\quad ou\quad x''=-8\end{array}$

Assim o conjunto solução é:

$\large\boxed{\begin{array}{l}\sf S=\Big\{-8~~;~~8\Big\}\end{array}}$

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G )

$\begin{array}{l}\sf 13x^2-5=6-2(2x^2-3)\\\\ \sf 13x^2-5=6-4x^2+6\\\\ \sf 13x^2+4x^2=6+6+5\\\\ \sf17x^2=17\\\\ \sf x^2=\dfrac{17}{17}\\\\ \sf x^2=1\\\\ \sf\sqrt{x^2}=\pm~\sqrt{1}\\\\ \sf x=\pm~1\\\\ \sf \therefore~~x'=1\quad ou\quad x''=-1\end{array}$

Assim o conjunto solução é:

$\large\boxed{\begin{array}{l}\sf S=\Big\{-1~~;~~1\Big\}\end{array}}$

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Att. Nasgovaskov

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