Math, asked by burhanuddinluqmani09, 1 year ago


 {x}^{2}  - 4 \sqrt{2} x + 6 = 0

Answers

Answered by Anonymous
0

Answer:

\boxed{x = \sqrt{2} , \bf{or}\: 3\sqrt{2}}

Step-by-step explanation:

Observe that:

6 = \sqrt{2}\times\sqrt{2}\times3

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x^2 - 4\sqrt{2} x + 6 = 0

\implies x^2 -3\sqrt{2} x -\sqrt{2} x +6 = 0

\implies x ( x - 3\sqrt{2} ) - \sqrt{2}( x - 3 \sqrt{2}) = 0

\implies (x - \sqrt{2} ) ( x - 3\sqrt{2} ) = 0

Two cases arise :-

\implies x = \sqrt{2}

\implies x = 3 \sqrt{2}


Hope it helps you

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