Question
Solve the following equation : $x ^2-4 \sqrt{2} x +6=0$
✓
Answer
$x^2-4 \sqrt{2} x+6=0$
$ x^2-\sqrt{2} x-3 \sqrt{2} x+6=0$
$ x(x-\sqrt{2})-3 \sqrt{2}(x-\sqrt{2})=0$
$(x-\sqrt{2})(x-3 \sqrt{2})=0 $
$ (x-\sqrt{2})=0,(x-3 \sqrt{2})=0 $
$ x=\sqrt{2}, x=3 \sqrt{2}$
$ x^2-\sqrt{2} x-3 \sqrt{2} x+6=0$
$ x(x-\sqrt{2})-3 \sqrt{2}(x-\sqrt{2})=0$
$(x-\sqrt{2})(x-3 \sqrt{2})=0 $
$ (x-\sqrt{2})=0,(x-3 \sqrt{2})=0 $
$ x=\sqrt{2}, x=3 \sqrt{2}$
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