Quadratic Equation - Sum and Product of Roots Cubic, Quadratic Equation

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Quadratic Equation: Sum and Product of Roots:

A general quadratic equation is given by

\begin{array}{l} a x^{2} + b x + c = 0 \end{array}

, where

\begin{array}{l} a, b \end{array}

and

\begin{array}{l} c \end{array}

are constants with

\begin{array}{l} a \neq 0 \end{array}

. The solutions or the roots of the above quadratic equation can be given by quadratic formula as :

\begin{array}{l} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{array}

\begin{array}{l} \Rightarrow x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \end{array}

The term

\begin{array}{l} b^{2} - 4 a c \end{array}

is the discriminant of the equation

\begin{array}{l} a x^{2} + b x + c = 0 \end{array}

and is denoted by D or Δ.

\begin{array}{l} \Rightarrow x = - \frac{b}{2 a} \pm \frac{\sqrt{D}}{2 a} \end{array}

If the roots of this equation are given by

\begin{array}{l} α \end{array}

and

\begin{array}{l} β \end{array}

, then there exists a relationship between the roots of quadratic equation and the coefficients

\begin{array}{l} a, b \end{array}

and

\begin{array}{l} c \end{array}

. The relationship is given as:

\begin{array}{l} α + β = - \frac{b}{a} \end{array}

\begin{array}{l} α β = \frac{c}{a} \end{array}

Sum of Roots

Thus, the sum of roots of a quadratic equation is given by the negative ratio of coefficient of

\begin{array}{l} x \end{array}

and

\begin{array}{l} x^{2} \end{array}

. The product of roots is given by ratio of the constant term and the coefficient of

\begin{array}{l} x^{2} \end{array}

.

Let us try to prove this graphically.

We know that the graph of a quadratic function is represented using a parabola. If α and β are the real roots of a quadratic equation, then the point of intersection of the plot of this function with x-axis represents its roots.

Let us assume that

\begin{array}{l} \frac{\sqrt{D}}{2 a} = n \end{array}

. Roots can be also given as

\begin{array}{l} α = - \frac{b}{2 a - n} \end{array}

—(1)

\begin{array}{l} β = - \frac{b}{2 a + n} \end{array}

—(2)

Graphically, it can be shown as:

Adding equations (1) and (2), we get

\begin{array}{l} α + β = - \frac{2 b}{2 a} \end{array}

\begin{array}{l} \Rightarrow α + β = - \frac{b}{a} \end{array}

Also,

\begin{array}{l} α = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{array}

\begin{array}{l} β = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} \end{array}

Multiplying both these equations we get

\begin{array}{l} α β = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \times \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} \end{array}

\begin{array}{l} \Rightarrow α β = \frac{(- b)^{2} - (b^{2} - 4 a c)}{4 a^{2}} \end{array}

\begin{array}{l} \Rightarrow α β = \frac{c}{a} \end{array}

Consider the general form of quadratic equation

\begin{array}{l} a x^{2} + b x + c = 0 \end{array}

.

Dividing both sides by

\begin{array}{l} ‘ a^{'} \end{array}

, we get,

\begin{array}{l} \frac{a}{a} x^{2} + \frac{b}{a} x + \frac{c}{a} = 0 \end{array}

\begin{array}{l} \Rightarrow x^{2} + \frac{b}{a} x + \frac{c}{a} = 0 \end{array}

Since,

\begin{array}{l} α + β = - \frac{b}{a} \end{array}

\begin{array}{l} α β = \frac{c}{a} \end{array}

\begin{array}{l} \Rightarrow x^{2} - (α + β) x + α β = 0 \end{array}

Thus by using roots, the equation can be obtained.

Consider the following two equations:

\begin{array}{l} x^{2} + 2 x - 3 = 0 \end{array}

and

\begin{array}{l} 2 x^{2} + 4 x - 6 = 0 \end{array}

.

The roots of these equations are equal i.e. 1 and -3. Graphically, these equations intersect thex-axis at the same points as shown below.

If the coefficients of two quadratic equations having same roots is given by

\begin{array}{l} a_{1}, b_{1}, c_{1} \end{array}

and

\begin{array}{l} a_{2}, b_{2}, c_{2} \end{array}

. Then, the following relation holds true

\begin{array}{l} \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}

The relationship between roots and coefficients of an equation can be established for higher order equations too. We know that the degree specifies how many roots an equation can have. The number of roots of an equation is equal to its degree. Let us take an

\begin{array}{l} n^{t h} \end{array}

order function and see how roots and coefficients are related to each other.

For a function

\begin{array}{l} f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots . . + a_{1} x + a_{0} \end{array}

, if the roots are given by

\begin{array}{l} α_{n - 1}, α_{n - 2} \dots \dots α_{0} \end{array}

,

then,

\begin{array}{l} f (x) = α_{n} (x - α_{n - 1}) (x - α_{n - 2}) \dots \dots \dots (x - α_{0}) \end{array}

\begin{array}{l} \Rightarrow a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots . . + a_{1} x + a_{0} = α_{n} (x - α_{n - 1}) (x - α_{n - 2}) \dots \dots \dots (x - α_{0}) \end{array}

On comparing coefficients of

\begin{array}{l} x^{n - 1} \end{array}

,we get:

\begin{array}{l} S_{1} = α_{n - 1} + α_{n - 2} + \dots + α_{0} = \sum_{i = 0}^{n - 1} α_{i} = - \frac{a_{n - 1}}{a_{n}} = (- 1)^{1} \end{array}

\begin{array}{l} \frac{c o e f f i c i e n t o f x^{n - 1}}{c o e f f i c i e n t o f x^{n}} \end{array}

On comparing coefficients of

\begin{array}{l} x^{n - 2} \end{array}

, we get:

\begin{array}{l} S_{2} = (α_{n - 1}) (α_{n - 2}) + (α_{n - 2}) (α_{n - 3}) + \dots = \sum_{i \neq j} α_{i} α_{j} = \frac{a_{n - 2}}{a_{n}} = (- 1)^{2} \frac{c o e f f i c i e n t o f x^{n - 2}}{c o e f f i c i e n t o f x^{n}} \end{array}

On comparing coefficients of

\begin{array}{l} x^{n - 3} \end{array}

, we get:

\begin{array}{l} S_{2} = (α_{n - 1}) (α_{n - 2}) (α_{n - 3}) + (α_{n - 2}) (α_{n - 3}) (α_{n - 4}) + \dots = \sum_{i \neq j \neq k} α_{i} α_{j} α_{k} \end{array}

=

\begin{array}{l} - \frac{a_{n - 3}}{a_{n}} = (- 1)^{3} \frac{c o e f f i c i e n t o f x^{n - 3}}{c o e f f i c i e n t o f x^{n}} \end{array}

Similarly,

\begin{array}{l} S_{n} = (α_{n - 1}) (α_{n - 2}) (α_{n - 3}) \dots α_{0} = (- 1)^{n} \frac{a_{0}}{a_{n}} = (- 1)^{n} \frac{c o n s t a n t t e r m}{c o e f f i c i e n t o f x^{n}} \end{array}

Thus, for a quadratic equation of the form

\begin{array}{l} a x^{2} + b x + c = 0 \end{array}

, having roots

\begin{array}{l} α \end{array}

and

\begin{array}{l} β \end{array}

,

\begin{array}{l} α + β = - \frac{b}{a} \end{array}

\begin{array}{l} α β = \frac{c}{a} \end{array}

For cubic equation of the form

\begin{array}{l} a x^{3} + b x^{2} + c x + d = 0 \end{array}

, having roots α,β and γ,

\begin{array}{l} α + β + γ = - \frac{b}{a} \end{array}

\begin{array}{l} α β + β γ + γ α = \frac{c}{a} \end{array}

\begin{array}{l} α β γ = - \frac{d}{a} \end{array}

For a fourth degree equation known as quartic equation in general form is given by:

\begin{array}{l} a x^{4} + b x^{3} + c x^{2} + d x + e = 0 \end{array}

If its roots are

\begin{array}{l} α, β, γ \end{array}

and

\begin{array}{l} δ \end{array}

then

\begin{array}{l} α + β + γ + δ = - \frac{b}{a} \end{array}

\begin{array}{l} α β + α γ + α δ + β γ + γ δ + δ β = \frac{c}{a} \end{array}

\begin{array}{l} α β γ + β γ δ + γ δ α + α β δ = - \frac{d}{a} \end{array}

\begin{array}{l} α β γ δ = \frac{e}{a} \end{array}

Thus, this process can be continued to find out the relationship between the roots and the coefficients of any equation of

\begin{array}{l} n^{t h} \end{array}

degree. Thus, sum and product of roots can be found out easily.’

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