Answer :

Now in the above quadratic equation the coefficient of x^{2} is 4. Let us make it unity by dividing the entire quadratic equation by 4.

x^{2} – 3/4x + 5/4 = 0

x^{2} – 3/4x = -5/4

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

Coefficient of x = 3/4

Half of 3/4 = 3/8

Squaring the half of 3/4 = 9/64

Now the LHS term is a perfect square and can be expressed in the form of (a-b) ^{2} = a^{2} – 2ab + b^{2} where a = x and b = 3/8

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

(x ± A)^{2} = k^{2}

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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