Question

factorize x^-2√5x+3 it

Answer 1

Heya mate
here is your answer

x^2-2√5x+3

=>(x-√5)^2 - 2

=>(x-√5)^2 - (√2)^2

[√2 ×√2=2]

By using a^2-b^2=(a+b) (a-b)

=>(x-√5+√2) (x-√5-√2)

I hope this will help you....

Answer 2

$= > {x}^{2} - 2 \sqrt{5} x + 3 \: \: \: \: \: \: [GIVEN]$
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◢HERE

$= > [a = 1 ,\: \: \: \: b = - 2 \sqrt{5}, \: \: \: \: c = 3]$

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●FACTORIZE FORMULA :-

$= > x = \frac{ - b \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ = > x = \frac{ - ( - 2 \sqrt{5} ) \binom{ + }{ - } \sqrt{ {( - 2 \sqrt{5}) }^{2} - 4 \times 1 \times 3 } }{2 \times 1} \\ \\ = > x = \frac{2 \sqrt{5} \binom{ + }{ - } \sqrt{20 - 12} }{2} \\ \\ = > x = \frac{2 \sqrt{5} \binom{ + }{ - } \sqrt{8} }{2} \\ \\ = > x = \frac{2 \sqrt{5} \binom{ + }{ - }2 \sqrt{2} }{2} \\ \\ = > x = \sqrt{5} \binom{ + }{ - } \sqrt{2} \: \: \: \: \: ....[Eq _{1}]$
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●SO BY Eq(1)

$= > x = \sqrt{5} + \sqrt{2} \\ \\ OR \\ \\ = > x = \sqrt{5} - \sqrt{2}$

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◢THEN,

$= > x = \sqrt{5} + \sqrt{2} \\ \\ = > [x - ( \sqrt{5} + \sqrt{2}) ]= 0 \\ \\ = > x - \sqrt{5} - \sqrt{2} = 0$

◢OR

$= > x = \sqrt{5} - \sqrt{2} \\ \\ = > x - \sqrt{5} + \sqrt{2} = 0$
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●THEN THE FACTOR OF GIVEN EQUATION IS

$= > {x}^{2} - 2 \sqrt{5} x + 3 \\ \\ = > (x - \sqrt{5} - \sqrt{2} )(x - \sqrt{5} + \sqrt{2} )$
____________________[◢ANSWER]

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