Waste less time on Facebook — follow Brilliant.
×

Quadratic equation

Hello, I have a misconception about the quadratics. I dont know whether i m right or not. Is x - (1/x) = 0 a quadratic or not? Some books say that the above one is a quadratic because the definition is "An algebraic equation of the form ax^2 + bx + c = 0 is a quadratic equation." But the graph of the given equation is not a parabola. So how can we say that the given one is a quadratic? Please clear my doubt.

Note by Shubham Poddar
1 week, 3 days ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

It depends.

The reason why people say it is a quadratic is that it is equivalent to a quadratic after you multiply by \(x\).

But otherwise, by the definition of "a quadratic equation is \( ax^2 + bx + c =0 \)", then no it is not a quadratic.

Calvin Lin Staff - 1 week, 3 days ago

Log in to reply

Is it necessary that graph of a quadratic is parabola?

Shubham Poddar - 1 week, 1 day ago

Log in to reply

Every particular class of second degree double variable equation (called a conic section) is represented by a family of curves of the same type. For example, \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) being a general ellipse and so on. The standard equation of all classes of conic sections are,

\[ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0\]

where letters \(a\) to \(f\) are constants that govern the class of the curve. For specifically parabolas, you have the term 'eccentricity' i.e., the value of \(h^2 - ab\) equal to 1.

Now the quadratic function is a special class of a parabolic conic section in which the axis of the parabola is parallel to the variable having degree 1 while the other variable should have a degree two like

\[ (y-k) = A(x-h)^2 + B(x-h) + C\]

where \(A,B,C\) and \(h,k\) are constants.

In your problem it is easy to see that if you multiply both sides with \(x\) then you get the quadratic equivalent of the equation. Notice, however that now the local maxima of the function on the LHS is defined and also the value of the function on the LHS at \(x=0\) is defined. However, that is, if the function on one side is treated in comparison to the original function; the equation is still satisfied by the same values for \(x\).

It is thus safe to say that the equation can be made quadratic but the function on the LHS is definitely not a quadratic function and thus the graph is a hyperbola and not a parabola.

Tapas Mazumdar - 4 days, 19 hours ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...