In a given quadratic equation $f(x)=ax^2+bx+c$ if $f(-1)>-4, f(1)<0$ and $f(3)>5$, then how can I find the sign of $a$?
Answer in the textbook: $a>0$
Finding sign of leading coefficient of a quadratic equation
quadratics
Related Solutions
1. what are symmetric expressions...
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 b if 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.
...in Quadratic Equations.
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 $\,q=\alpha\beta\,$.
2. Why they are generally expressed in terms of $\alpha + \beta$ and $\alpha \beta$?
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.
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?
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 $\,\alpha,\beta\,$.
Divide your equation by $Q2^2$. You will have a quadratic in $\frac {Q1}{Q2}$ which completing the square or the quadratic formula will solve.
Because of the symmetry of the problem some ratio for $\frac {Q1}{Q2}$ and its inverse must both be solutions because you can rename the two spheres and get the reciprocal ratio. Only choice c gives two choices, so it has to be right. You can verify $-(3+\sqrt 8)(-3+\sqrt 8)=1$ as a check.
Best Answer
If $a=0$, then you have a straight line. Then $$f(1)=\frac{f(-1)+f(3)}2>\frac{-4+5}2=\frac 12>0$$ Then for a parabola with $a>0$ you have $$f(\frac{x+y}2)<\frac{f(x)+f(y)}2$$ and for $a<0$ you have $$f(\frac{x+y}2)>\frac{f(x)+f(y)}2$$ Since $$\frac{f(-1)+f(3)}2>0$$ and $$f(1)<0$$ then $$f(\frac{-1+3}2)<\frac{f(-1)+f(3)}2$$