√2x²+7x+5√2=0 solve by factorization
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21
Hi Friend !!!
Here is ur answer !!!
√2x²+7x+5√2 = 0
√2x²+2x+5x+5√2 = 0
√2x(x+√2) + 5(x+√2) = 0
(√2x+5)(x+√2) = 0
x = -5/√2 ----OR ------√2
Hope it helps u : )
Here is ur answer !!!
√2x²+7x+5√2 = 0
√2x²+2x+5x+5√2 = 0
√2x(x+√2) + 5(x+√2) = 0
(√2x+5)(x+√2) = 0
x = -5/√2 ----OR ------√2
Hope it helps u : )
Answered by
6
=√2 x^2 +7x + 5√2=0
product of coefficient of x^2 and constant term is 10 so break 7 in such parts that it's sum is 7 but product is 10 so
=√2 x^2 +2x+5x +5√2=0
=√2x(x+√2) + 5(x+√2)
so
(√2x+5)(x+√2)
so x = -5/√2 , -√2
hope it helped you please mark as brainliest
product of coefficient of x^2 and constant term is 10 so break 7 in such parts that it's sum is 7 but product is 10 so
=√2 x^2 +2x+5x +5√2=0
=√2x(x+√2) + 5(x+√2)
so
(√2x+5)(x+√2)
so x = -5/√2 , -√2
hope it helped you please mark as brainliest
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