Question

Solve the quadratic equations by Factorization method :
$\sf{x^{2} + 2 \sqrt{2x} - 6 = 0 }$
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Answer 1

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$\sf \underline{\red{ \: \: \: \: \: \: \: \: \: \: \: \: \: Aɴsᴡᴇʀ \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}$

$\scriptsize \sf{ We \: have \: , \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \scriptsize \sf{x ^{2} + 2 \sqrt{2x} - 6 = 0 } \\ \scriptsize \sf{ \implies \: x^{2} + 3 \sqrt{2} x - \sqrt{2x} - 6 = 0 } \: \: \: \: \: \: \: \: \: \\ \scriptsize \sf{ \implies \: x(x + 3 \sqrt{2}) - \sqrt{2}(x + 3 \sqrt{2}) = 0 } \\ \scriptsize \sf{ \implies \: (x + 3 \sqrt{2})(x - \sqrt{2}) = 0 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \scriptsize \sf{ \implies \: x + 3 \sqrt{2} = 0 \: or , \: x - \sqrt{2} = 0 } \: \: \: \: \: \\ \scriptsize \sf{ \implies \: x = - 3 \sqrt{2} \: or \: , \: x = \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$$\scriptsize \sf{Thus , x = - 3 \sqrt{2} \: and \: x = \sqrt{2} \: are \: two \: roots \: of \: the \: given \: equation . }$

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Answer 2

Answer:

see the attachment above.

hope it helpss you..