Question

Verify (x+1/x)^2=3(x+1/x)+4 equation is quadratic or not ? Explain

Answer 1

The equation is (x+1/x)^2=3(x+1/x)+4

Then L.H.S. becomes,
x^2+2x+1/x^2

and the R.H.S. becomes,
3x+3+4x/x=7x+3/x

Simplifying both the sides,
x^2+2x+1/x=7x+3
=x^2+2x+1=x(7x+3)
=x^2+2x+1=7x^2+3x

=7x^2-x^2+3x-2x=1
=6x^2+x-1=0 is the required equation.

It is a quadratic equation because it is in the form of ax^2+bx+c=0 and the degree of polynomial is 2.

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Answer 2

Answer:

To verify whether the given equation (x+1/x)^2=3(x+1/x)+4 is quadratic or not, we need to check if the highest degree of the variable,

Expanding the left-hand side of the equation, we get:

(x+1/x)^2 = (x^2 + 2 + 1/x^2)

And expanding the right-hand side of the equation, we get:

3(x+1/x)+4 = 3x + 3/x + 4

Simplifying both sides, we get:

x^2 + 2 + 1/x^2 = 3x + 3/x + 4

Multiplying both sides by x^2, we get:

x^4 + 2x^2 + 1 = 3x^3 + 3x + 4x^2

Bringing all the terms to one side, we get:

x^4 - 3x^3 + 6x^2 - 3x + 1 = 0

Since the highest degree of x is 4, which is 2nd order or quadratic, we can conclude that the given equation (x+1/x)^2=3(x+1/x)+4 is quadratic.

Therefore, the given equation is quadratic.

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