Question

Could someone help us with the 2nd and the 3rd question plzzz it’s urgent.

Answer 1

Question No.(1) :-----›
_____________

Answer : NO

Solution :
________

Given that : The quadratic equation is
${x}^{2} - 4x + 3 \sqrt{2}$

we have to find that is x = √2 a root of this equation.

If x=√2 satisfy the equation then it will be a root of this equation

On putting x = √2 in equation

${ (\sqrt{2} )}^{2} - 4 \sqrt{2} + 3 \sqrt{2} = 0 \\ \\ = > 2 - \sqrt{2} = 0 \\ \\ LHS \: is \: not \: equal \: to \: RHS \:$

So, x = √2 is not a root of the equation.

Question No. (2) : -----›
_____________

Answer : 7

Solution :
________

Given that :

$( - x) \: (x - 3) \: and \: (x + 4)$

Are in A.P.

If they are in A.P. then,

$\frac{ (- x) + (x + 4)}{2} = (x - 3) \\ \\ = > \frac{4}{2} = (x - 3) \\ \\ = > x - 3 = 2 \\ \\ = > x = 7$

So, the value of x will be 7