Question
Answers

For a quadratic equation, if one of the root is $\sqrt 2 + \sqrt 3 $, then the equation is ${x^2} - 2\sqrt 2 x - 1 = 0$.
A. True
B. False

Answer Verified Verified
Hint:Use Shreedharacharya formula to calculate the roots of the quadratic equation. Then check whether one of the roots is$\sqrt 2 + \sqrt 3 $.

Formula used:Shreedharacharya formula for finding roots of quadratic equation. $ x = \dfrac{{ - b \pm \sqrt D }}{{2a}} = \dfrac{{ - ( - 2\sqrt 2 ) \pm \sqrt {{{( - 2\sqrt 2 )}^2} - 4(1)( - 1)} }}{{2(1)}}$

Complete step-by-step answer:
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.

So, the correct answer is “Option A”.

Note:While calculating for the roots in a quadratic equation $ax^2+bx+c=0$ we first need to find the D which is also called the discriminant i.e $D=b^2-4ac$. If this discriminant is more than zero then the roots of the equation are real.