Root 2 x square + 7x + 5root 2 =0
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7
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√2 x2 + 7x + 5√2 = 0
√2 x2 + 2x+ 5x + 5√2 = 0
√2x (x+√2 ) + 5 ( x + √2 ) = 0
( √2 + x ) ( √2 x + 5 ) =0
therefore x = -√2 , x = - 5 / √2
√2 x2 + 7x + 5√2 = 0
√2 x2 + 2x+ 5x + 5√2 = 0
√2x (x+√2 ) + 5 ( x + √2 ) = 0
( √2 + x ) ( √2 x + 5 ) =0
therefore x = -√2 , x = - 5 / √2
Answered by
1
rt2x^2 + 7x +5rt2 = (rt2x^2 + 2x) + (5x + 5rt2)
= rt2x ( 1x + rt2 ) + 5 (1x + rt2)
= (rt2x + 5) ( 1x + rt2) =0
Implies (rt2x + 5) =0 or (1x + rt2)=0
x=-5/rt2 or x=-rt2
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