What is Symmetric Expression in Quadratic Equation?
As per the definition of my textbook, –
The Symmetric Expressions of the roots $\alpha$, $\beta$ of an
equation are those expressions in $\alpha$ and $\beta$ which do not
change by interchanging a and b. To find the value of such expression,
we generally, express that in terms of $\alpha + \beta$ and $\alpha
> \beta$.
- $\alpha^2 + \beta^2$
- $\alpha^2 + \alpha \beta + \beta^2$
- $\frac 1 \alpha + \frac 1 \beta$
and this continues to some expressions.
So, I'm unable to understand –
1. what are symmetric expressions in Quadratic Equations. (Can anyone explain by Diagram, then that'd surely help)
2. Why they are generally expressed in terms of $\alpha + \beta$ and $\alpha \beta$?
3. The given examples of the symmetric expressions (as given above 1, 2 and 3) aren't quadratic equations really, they seem to like in 2 variable. What are they? How are they derived? and What can be done it?
Although I searched on Wiki and other websites on this topic, most of them were highly complicated beyond my level so, I can't seem to understand those
Best Answer
As noted in the first paragraph, symmetric expressions are
those expressions in α and β which do not change by interchanging α and β.An expression in $\,a,b\,$ can be written as $\,f(a,b)\,$. Interchanging $\,a\,$ and $\,b\,$ gives the expression $\,f(b,a)\,$. By definition, the original expression is
symmetric in a and bif and only if $\,f(a,b)=f(b,a)\,$ for all $\,a,b\,$.$\,f(a,b)=a^2+b^2+ab+1\,$ is symmetric in $\,a,b\,$ because $\,f(b,a)=b^2+a^2+ba+1$ $=a^2+b^2+ab+1=f(a,b)\,$ for all $\,a,b\,$.
$\,f(a,b)=a^2-2b^2\,$ is not symmetric in $\,a,b\,$ because $\,f(b,a)=b^2-2a^2 \ne a^2-2b^2=f(a,b)\,$. Even though it is possible that $\,f(a,b)=f(b,a)\,$ for some $\,a,b\,$ e.g. $\,f(1,-1)=f(-1,1)\,$, the equality does not hold true for all values e.g. $\,f(1,0)=1 \ne -2 = f(0,1)\,$, so the expression is not symmetric.
It's not about "symmetric expressions in quadratic equations", but rather "symmetric expressions in the roots of quadratic equations".
The first paragraph refers to
the roots α, β of an equation. Since there are two roots, and assuming the context is algebraic, the equation must be a quadratic, such as $\,(x-\alpha)(x-\beta)=0\,$. Expanding and collecting, the quadratic can also be written as $\,x^2-px+q=0\,$, where $\,p=\alpha+\beta\,$ and $\,q=\alpha\beta\,$.Because it is possible: by the fundamental theorem of symmetric polynomials, any polynomial expression in $\,\alpha,\beta\,$ which is symmetric in $\,\alpha,\beta\,$ can be expressed as a polynomial in the elementary symmetric polynomials, which in the case of two variables are just $\,\alpha+\beta\,$ and $\,\alpha \beta\,$.
Because it is convenient: by Vieta's formulas, the elementary symmetric polynomials in the roots of a polynomial are directly related to the coefficients of the equation. In the simple case of a quadratic $\,x^2-px+q=0\,$, Vieta's formulas are precisely $\,\alpha+\beta=p\,$ and $\,\alpha\beta=q\,$ from the previous step.
Those are symmetric expressions in the roots of a quadratic equation.
Taking the first one for example, the complete statement of the problem would be something like:
$$ \style{font-family:inherit}{\text{Find the value of the expression}} \;\alpha^2+\beta^2\; \style{font-family:inherit}{\text{where}} \;\alpha,\beta\; \style{font-family:inherit}{\text{are the roots of}} \;x^2 - px+q=0\,. $$
Using the hint, and using that $\,\alpha+\beta=p\,$ and $\,\alpha \beta=q\,$, the expression can be calculated as:
$$ \alpha^2+\beta^2\color{red}{+2\alpha\beta-2\alpha\beta} = (\alpha+\beta)^2 - 2 \alpha\beta = p^2 - 2q $$
It would be technically possible, of course, to actually solve the quadratic, find the roots $\,\alpha\,$ and $\,\beta\,$ explicitly, then calculate $\,\alpha^2+\beta^2\,$ the long and hard way. However, since the expression was symmetric, it was possible to calculate it a lot more quickly without having to actually find the individual values of $\,\alpha,\beta\,$.