Theory of Equations - JEE Important Topic

The Theory of equations is essentially the study of the inter-relationship between an equation and the polynomials that comprise the equation. An equation is basically a mathematical expression, equating two different quantities, values, or expressions. The main issue in this field of study is that an algebraic equation will have an algebraic solution. To put it simply, the idea of complex solutions was not common knowledge.

Therefore, finding the solution to a single unknown variable in a non-linear polynomial equation was a stumbling block. Mathematician Evariste Galois was the first to solve this problem in the year 1830. His theory gave a distinct outline to help understand which equations could be solved by radical values. This is how the theory of equations became its own area rather than being synonymous with algebra itself. 

Definition and Mathematical Theory of Equations

Theory of equations can be defined as the study of the different methods that can be implied to find out the unknown values and solve a mathematical equation.

Mathematical Theory of Equation:

A polynomial function or equation can be represented by the expression, $f(x)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+...+{{a}_{n}}$.

Here, n is a non-negative integer and ${{a}_{i}}(i=0,1,...,n)$ are fixed complex numbers. The above expression in f will be termed a polynomial of x up to the nth degree. ${{a}_{1}},{{a}_{2}},...,{{a}_{n}}$ are termed as coefficients of the function $f$.

This is the basic theory of equations behind the concept of the theory of equations.

In the case of a quadratic equation, the general formula is given by $a{{x}^{2}}+bx+c=0$. (Here a,b,c are arbitrary constants of the equation).

Every quadratic equation has two roots, which can be either real roots, or complex in nature. The discriminant of the quadratic equation is given by the expression, $D={{b}^{2}}-4ac$. The nature of the discriminant decides whether the equation will have two real roots, two equal roots or only complex roots. 

Key Points Regarding the Theory of Equations

To summarise the important concepts and ideas related to the theory of equations, the following list can be helpful:

• The general mathematical formula for a quadratic equation in variable x is given by $a{{x}^{2}}+bx+c=0$.

• The roots of a quadratic equation are given by $x=-b\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{2a}$

• If $\alpha ,\beta$ are the two roots of the equation, then $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \cdot\beta =\dfrac{c}{a}$

• The discriminant of the quadratic equation is denoted by $D={{b}^{2}}-4ac$.

• When D = 0, the equation has two equal roots.

• When D > 0, the equation has two distinct real roots.

• When D < 0, the roots of the equation will be complex. 

• The quadratic equation exhibits a parabolic graph which opens upwards if a > 0 and downwards if a < 0.

• At a > 0, the function f has the minimum value and when a < 0 the function has its maximum value.

Solved Problems on Theory of Equations 

To understand better how this math theory is exactly applied, we will take the following examples.

Problem 1: If the roots of the equation ${{x}^{3}}+p{{x}^{2}}+qx+r=0$ are in arithmetic progression, then prove that $2{{p}^{3}}-9pq+27r=0$

Solution:

Given equation, ${{x}^{3}}+p{{x}^{2}}+qx+r=0$. 

The roots of the equation are in arithmetic progression. Therefore, let the roots be a-d, a, a+d respectively.

$\begin{align} & \therefore (-p)=a-d+a+a+d \\  & \Rightarrow -p=3a \\  & \Rightarrow a=\dfrac{-p}{3} \\ \end{align}$

As a is one of the three roots of the given equation, it must satisfy the equation, ${{x}^{3}}+p{{x}^{2}}+qx+r=0$.

Therefore, substituting x with $a=\dfrac{-p}{3}$, we get:

${{\left( \dfrac{-p}{3} \right)}^{3}}+p{{\left( \dfrac{-p}{3} \right)}^{2}}+q\left( \dfrac{-p}{3} \right)+r=0$

On simplification, we get , $2{{p}^{3}}\,-9pq+27r=0$.  Hence, proved.

Problem 2: If $\alpha$ and $\beta$ are the roots of the quadratic equation $2{{x}^{2}}+8x+k=0$, what is the maximum value of $\left[ \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha } \right]$ when $k<0$.

Theory:

Comparing the quadratic equation, $2{{x}^{2}}+8x+k=0$ with $a{{x}^{2}}+bx+c=0$ we get,

$a=\,2,\,b=\,8$ and $c=k$

Let the roots of the given equation be $\alpha$ and $\beta$.

Then sum of roots, $\alpha +\beta =-\,\dfrac{b}{a}$, 

$\therefore \,\,\alpha +\beta =-\dfrac{8}{2}=-\,4$ and,

Product of roots is  $\alpha \bullet \beta =\,\dfrac{c}{a}$ Or $\alpha \bullet \beta =\dfrac{k}{2}$

Discriminant, $D={{b}^{2}}-\,4ac=64-8k$. 

Now since, k < 0, $\therefore d=64-8k<0$. 

As the discriminant is less than zero, the roots of the equation are both real roots.

Now, $\left[ \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha } \right]=\left[ \dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right]=\left[ \dfrac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta } \right]$

$\Rightarrow \left[ \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha } \right]=\left[ \dfrac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta } \right]=\left[ \dfrac{{{(\alpha +\beta )}^{2}}}{\alpha \beta }-2 \right]$

Substituting the values of $\alpha +\beta$ and $\alpha \beta$ in$\left[ \dfrac{{{(\alpha +\beta )}^{2}}}{\alpha \beta }-2 \right]$  we get,

$\left[ \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha } \right]\,\,=\,\,\left[ \dfrac{{{(-4)}^{2}}}{\left(\dfrac{k}{2}\right)}-2 \right]=\,\dfrac{32}{k}-\,2$ 

As, k < 0, therefore the maximum value of $\left[ \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha } \right]=-2$.

Conclusion

The theory of equations basically gives us the different methods that can be used to solve polynomial equations. A polynomial expression can either be linear or nonlinear. The general formula of a quadratic function in x is given by $a{{x}^{2}}+bx+c=0$ and the discriminant the discriminant is given by D, such that $D={{b}^{2}}-4ac$ . The discriminant decides the nature of the roots. The roots of a quadratic equation can either be equal, real or complex. The sum of the roots of the equation gives us the value of $\dfrac{-b}{a}$ and the product of the roots gives us the value of $\dfrac{c}{a}$.