Q. 1 F5.0( 1 Vote )

Solve the followi

Answer :

Now in the above quadratic equation the coefficient of x2 is 2. Let us make it unity by dividing the entire quadratic equation by 2.

x2 + 1/2 x + 2= 0


x2 + 1/2 x = - 2


Now by taking half of the coefficient of x and then squaring it and adding on both LHS and RHS sides.


Coefficient of x = 1/2


Half of 1/2 = 1/4


Squaring the half of 1/2 = 1/16



Now the LHS term is a perfect square and can be expressed in the form of (a-b) 2 = a2 – 2ab + b2 where a = x and b = 1/4



On simplifying both RHS and LHS we get an equation of following form,


(x ± A)2 = k2



It is observed that the term obtained on RHS is a negative term and taking square root of a negative term will give imaginary roots for the given quadratic equation.


Therefore the given quadratic equation does not has real roots.


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