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A. True

B. False

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We are given that a quadratic ${x^2} - 2\sqrt 2 x - 1 = 0$ has one of the roots $\sqrt 2 + \sqrt 3 $. We need to check whether the statement is true or not.

To find the roots of the equation we need to use the Shreedharacharya formula. We will start from understanding the Shreedharacharya formula and it’s derivation.

Let the general quadratic equation be $a{x^2} + bx + c = 0$. We have,

$a{x^2} + bx + c = 0$

Multiply both sides by 4a

$ \Rightarrow 4{a^2}{x^2} + 4abx + 4ac = 0$

Subtracting both side by 4ac

$ \Rightarrow 4{a^2}{x^2} + 4abx = - 4ac$

Adding both sides ${b^2}$ we get,

$ \Rightarrow 4{a^2}{x^2} + 4abx + {b^2} = - 4ac + {b^2}$

Completing the square on LHS we get,

$ \Rightarrow {(2ax + b)^2} = - 4ac + {b^2} = D$

Taking square roots we get,

$ \Rightarrow 2ax + b = \pm \sqrt D .$

Subtracting both sides b and dividing by 2a we get,

$ \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}}.…………. (1)$

Thus roots of quadratic equation are given by $ \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}}.$

Now, using the Shreedharacharya formula in ${x^2} - 2\sqrt 2 x - 1 = 0$.

$a = 1$

$b = - 2\sqrt 2 $

$c = - 1$

Putting the values of a, b and c in equation (1) we get,

\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}} = \dfrac{{ - ( - 2\sqrt 2 ) \pm \sqrt {{{( - 2\sqrt 2 )}^2} - 4(1)( - 1)} }}{{2(1)}}\].

Simplifying the values we get,

\[ \Rightarrow x = \dfrac{{2\sqrt 2 \pm \sqrt {8 + 4} }}{2} = \dfrac{{2\sqrt 2 \pm \sqrt {12} }}{2}\]

Calculating it further we have,

\[ \Rightarrow x = \dfrac{{2\sqrt 2 \pm 2\sqrt 3 }}{2}\]

Taking 2 common in numerator and cancelling out on the RHS we have,

\[ \Rightarrow x = \sqrt 2 \pm \sqrt 3 \]

This shows that the quadratic equation has roots $\sqrt 2 + \sqrt 3 $ and $\sqrt 2 - \sqrt 3 $.

Since, $\sqrt 2 + \sqrt 3 $ is a root of the quadratic equation ${x^2} - 2\sqrt 2 x - 1 = 0$. Therefore, the statement given is true.

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