Question

If one root of the equation is √3 + 2, form the equation.
(a)x² - 2√3 – 1=0
(b)x²-3x+1=0
(C)x²-5x+1=0
(d)x² - 4x+1=0
​

Answer 1

One root of the equation is given as $\sf \sqrt{3} + 2$

And we are asked to find the equation.

The concept that we are supposed to apply here is that :-

irrational roots always found in conjugate pairs.

hence, the second root of this equation will be :-

$\longrightarrow \sf\sqrt{3} - 2$

Now, we can easily find the equation by using the following formula:-

$\large \underline {\boxed {\pink{\frak{equation = {x}^{2} - (sum \: of \: roots)x + (product \: of \: roots)}}}} \bigstar$

Let's calculate the sum of roots :-

$: \implies \sf \sqrt{3} + 2 + \sqrt{3} - 2$

$: \implies \boxed {\pink{\frak 2 \sqrt{3}} } \bigstar$

Now, calculate the product of roots :-

$: \implies \sf ( \sqrt{3 } + 2)( \sqrt{3} - 2 )$

$: \implies \sf( { \sqrt{3} })^{2} - ( {2})^{2}$

$: \implies \sf 3 - 4$

$: \implies{ \boxed{ \pink{{ \frak{ - 1 }}}}} \bigstar$

$\dag \: \underline{ \frak{substituting \: these \: values \: in \: formula : - }}$

$: \implies \sf p(x) = {x}^{2} - 2 \sqrt{3} x + ( - 1)$

$: \implies \boxed {\pink{{ \frak{ p(x) = {x}^{2} - 2 \sqrt{3} x - 1}} }} \bigstar$

$\therefore \sf\underline{option \: \bold{A} \: is \: our \: right \: answer}$