Q. 1 D

Solve the followi

Now in the above quadratic equation the coefficient of x2 is 4. Let us make it unity by dividing the entire quadratic equation by 4.

x2 + √3x + 3/4 = 0

x2 + √3x = -3/4

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

Half of √3= √3/2

Squaring the half of √3= 3/4

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 = √3/2

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

(x ± A)2 = k2

Here RHS term is zero which implies that the roots of the quadratic equation are real and equal.

Taking square root of both sides,

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