Question

if 2 minus root 3 is a root of quadratic equation X square + 2 ( root3 - 1) X + 3 - 2 root 3 equals to zero then the second root is​

Answer 1

Answer :-

$\implies \boxed{ \sf{ \beta = - \sqrt{3} }} \: \\ \: \leadsto \bf{ other \: root \: is \: - \sqrt{3} }$

To Find :-

→ Other root of given quadratic equation .

Explanation :-

According to the question ;

$\to \sf{ 2 - \sqrt{3} \: \: is \: a \: root \: of \: } \\ \to \sf{ {x}^{2} + 2( \sqrt{3} - 1)x + 3 - \sqrt{3} = 0} \: \\ \\ \sf{ let \: \: \: \alpha \: \: and \: \beta \: are \: the \: roots \: of \:} \\ \sf{ given \: equation \: \: }$

We know that ;

$\to \boxed{\sf{ sum \: of \: roots = \frac{ - cofficient \: of \: x}{ coefficient \: of \: {x}^{2} } }} \\ \\ \therefore \: \\ \implies \sf{ \: \alpha + \beta = \frac{ - 2( \sqrt{3} - 1) }{1} } \\ \\ \because \boxed{\sf{ \alpha = 2 - \sqrt{3} } \: \: (given \: root)} \\ \\ \therefore \: \\ \implies \: 2 - \sqrt{3} + \beta = - 2 \sqrt{3} + 2 \\ \\ \implies \: \beta = 2 - 2 \sqrt{3} - 2 + \sqrt{3} \\ \\ \implies \boxed{ \sf{ \beta = - \sqrt{3} }}$

Hence other root of given quadratic equation is -√3 .

Verification :-

Replace all x to -√3

$\to \: {( - \sqrt{3}) }^{2} + 2( \sqrt{3} - 1)( - \sqrt{3} ) + 3 - 2 \sqrt{3} = 0 \\ \\ \to \: 3 - (2 \sqrt{3} - 2)( \sqrt{3} ) + 3 - 2 \sqrt{3} = 0 \\ \\ \to \: 3 - 6 + 2 \sqrt{3} + 3 - 2 \sqrt{3} = 0 \\ \\ \to \: 6 - 6 = 0 \\ \\ \to \: 0 = 0$

Hence verified .

Answer 2

Hii guys ❣️❣️

Step-by-step explanation:

First root is given in question - 2-√3

We have to find second root - ?

So, we can say α = 2-√3

And

β = ?

I hope this is helpful for you

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