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MissTica
3 years ago
10

Find the root of

rt%7B2%7D%20%20%3D%200" id="TexFormula1" title=" \sqrt{2x { }^{2} } + x + \sqrt{2} = 0" alt=" \sqrt{2x { }^{2} } + x + \sqrt{2} = 0" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
artcher [175]3 years ago
6 0

Answer:

The root of    x = -2 + √2

Step-by-step explanation:

<u><em> Step(i):-</em></u>

Given \sqrt{2x^{2} } + x +\sqrt{2} = 0

      ⇒    \sqrt{2} } ( x^2 ) ^{\frac{1}{2} } + x +\sqrt{2} = 0

      ⇒   \sqrt{2}x  + x +\sqrt{2} = 0

     ⇒      (\sqrt{2}  + 1) x = - \sqrt{2}

           x = \frac{-\sqrt{2} }{(\sqrt{2}  + 1)}

<u><em>Step(ii):-</em></u>

Rationalizing

     x = \frac{-\sqrt{2} }{(\sqrt{2}  + 1)} X \frac{(\sqrt{2}  - 1)}{(\sqrt{2}  - 1)}

Apply ( a-b ) ( a+b) = a²-b²

    x = \frac{-\sqrt{2}(\sqrt{2} -1) }{(\sqrt{2})^{2}   - 1^{2} )} = \frac{-(2-\sqrt{2} }{2-1}

The root of    x = -2 + √2

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