Q. 1 A4.0( 6 Votes )

Solve the followi

Answer :

Now in the above quadratic equation the coefficient of x2 is 3. Let us make it unity by dividing the entire quadratic equation by 3.

x2 – 5/3x + 2/3 = 0

x2 – 5/3x = -2/3

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

Coefficient of x = 5/3

Half of 5/3 = 5/6

Squaring the half of 5/3 = 25/36

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 = 5/6

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

(x ± A)2 = k2

Taking Square root of both sides.

Now taking the positive part,

x = 6/6

x = 1

Now taking the negative part,

x = 4/6

x = 2/3

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