Question

please solve it please​

Answer 1

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$2x {}^{2} - 2 \sqrt{2} x - 3 = 0$

$d = b {}^{2} - 4ac$

$d = ( - 2 \sqrt{2} ) {}^{2} - 4 \times 2 \times ( - 3)$

$d = 8 + 24$

$d = 32$

Now,

$x = \frac{( - b) + \sqrt{d} }{2a}$

$x = \frac{ - ( - 2 \sqrt{2}) + 32 }{2 \times 2}$

$x = \frac{2 \sqrt{2} + 32 }{4}$

$x = \frac{2( \sqrt{2} + 16)}{4}$

$x = \frac{ \sqrt{2} + 16 }{2}$

And,

$x = \frac{( - b) - \sqrt{d} }{2a}$

$x = \frac{ - ( - 2 \sqrt{2}) - 32 }{2 \times 2}$

$x = \frac{2 \sqrt{2} - 32 }{4}$

$x = \frac{2( \sqrt{2} - 16)}{4}$

$x = \frac{ \sqrt{2} - 16 }{2}$

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Answer 2

Answer:

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