Question

x²-(root 3+1)x+root 3=0 by completing square method​

Answer 1

Answer:

√3 and 1

Step-by-step explanation:

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$\sf x {}^{2} - ( \sqrt{3} + 1)x + \sqrt{3} = 0 \\ \\ \sf \rightarrow x {}^{2} - \frac{2}{2}.( \sqrt{3} + 1)x + \sqrt{3} = 0 \\ \\ \sf \rightarrow x {}^{2} - 2 \: . \: x \: . \: \frac{ \sqrt{3} + 1}{2} + \sqrt{3} = 0 \\ \\ \sf \rightarrow \tiny x {}^{2} - 2 \: . \: x \: . \: \frac{ \sqrt{3} + 1}{2} + ( \frac{ \sqrt{3} + 1 }{2} ) {}^{2} + \sqrt{3} = ( \frac{ \sqrt{3} + 1}{2} ) {}^{2} \\ \\ \sf \rightarrow \{x - ( \frac{ \sqrt{3} + 1}{2} ) \} {}^{2} + \sqrt{3} = (\frac{ \sqrt{3} + 1}{2} ) {}^{2} \\ \\ \sf \rightarrow \{x - ( \frac{ \sqrt{3} + 1}{2} ) \} {}^{2} = \{(\frac{ \sqrt{3} + 1}{2} ) {}^{2} - \sqrt{3} \} \\\\ \rightarrow \sf \{x - ( \frac{ \sqrt{3} + 1 }{2} ) \} {}^{2} = \frac{3 + 1 + 2 \sqrt{3} }{4} - \sqrt{3} \\ \\ \rightarrow \sf \{x - ( \frac{ \sqrt{3} + 1 }{2} ) \} {}^{2} = \frac{3 + 1 + 2 \sqrt{3} - 4 \sqrt{3} }{4} \\ \\ \rightarrow \sf \{x - ( \frac{ \sqrt{3} + 1 }{2} ) \} {}^{2} = \frac{3 + 1 - 2 \sqrt{3} }{4}\\ \\ \sf \rightarrow \{x - ( \frac{ \sqrt{3} + 1}{2} ) \} {}^{2} = \{(\frac{ \sqrt{3} - 1}{2} ) {}^{2} \}\\ \\ \bf on \: cancelling \: square \: both \: sides - \\ \\ \sf \rightarrow \{x - ( \frac{ \sqrt{3} + 1}{2} ) \} = \pm(\frac{ \sqrt{3} + 1}{2} ) \\ \\ \sf \rightarrow x = \frac{ \sqrt{3} + 1}{2} \pm \frac{ \sqrt{3} - 1}{2}$

Therefore, roots of the equation are:

$\sf \rightarrow x = \frac{ \sqrt{3} + 1 }{2} + \frac{ \sqrt{3} - 1}{2} \\ \\ \sf \rightarrow x = \frac{ \sqrt{3} + 1 + \sqrt{3}- 1}{2} \\ \\ \sf \rightarrow x = \frac{2 \sqrt{3} }{2} \\ \\ \boxed{\sf \therefore \bf x = \sqrt{3} }$

Also,

$\sf \rightarrow x = \frac{ \sqrt{3} + 1}{2} - ( \frac{ \sqrt{3} - 1}{2} ) \\ \\ \sf \rightarrow x = \frac{ \sqrt{3} + 1 - \sqrt{3} + 1}{2} \\ \\ \sf \rightarrow x = \frac{2}{2} \\ \\ \boxed{ \bf \therefore x = 1}$

Hence, it's roots are √3 and 1.

Answer 2

hope it helps

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