Equation of the form $a x^2+b x+c=0$ where $a, b, c \in R$ and $a \neq 0$ is called a quadratic equation, where $a, b, c$ are coefficients of this equation.

first divide it by $a$.

we get $x^2+\frac{b}{a} x+\frac{c}{a}=0$

$$ \begin{aligned} &\Rightarrow \left(x+\frac{b}{2 a}\right)^2+\frac{c}{a}-\frac{b^2}{4 a^2}=0 \\\\ &\Rightarrow \left(x+\frac{b}{2 a}\right)^2=\left(\frac{b^2-4 a c}{4 a^2}\right) \\\\ &\Rightarrow x+\frac{b}{2 a}=\pm \sqrt{\frac{b^2-4 a c}{4 a^2}} \\\\ &\Rightarrow x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \end{aligned} $$

For the quadratic equation $a x^2+b x+c=0(a \neq 0)$ has two roots, given by

$$ \alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a} \text { and } \beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a} $$

NOTE :

(a) Every quadratic equation has two and only two roots.

(b) The expression $b^2-4 a c=D$ is called the discriminant of the quadratic equation.

(c) If $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^2+b x+c=0$, then:

$\begin{array}{lll}\text { (i) } \alpha+\beta=-b / a & \text { (ii) } \alpha \beta=c / a & \text { (iii) }|\alpha-\beta|=\sqrt{D} /|a|\end{array}$

(d) A quadratic equation whose roots are $\alpha$ and $\beta$ is $(x-\alpha)(x-\beta)=0$

i.e., $x^2-(\alpha+\beta) x+\alpha \beta=0$

$\Rightarrow x^2-$ (sum of roots) $x+$ product of roots $=0$.

If roots of quadratic equation $a x^2+b x+c=0(a \neq 0)$ are $\alpha$ and $\beta$ then

(i) $|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4 \alpha \beta}=\left|\frac{\sqrt{b^2-4 a c}}{a}\right|=\left|\frac{\sqrt{D}}{a}\right|$

(ii) $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=\frac{b^2-2 a c}{a^2}$

(iii) $\left|\alpha^2-\beta^2\right|=|\alpha+\beta| \sqrt{(\alpha+\beta)^2-4 \alpha \beta}=\left|\frac{b \sqrt{b^2-4 a c}}{a^2}\right|$

(iv) $\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)=-\frac{b\left(b^2-3 a c\right)}{a^3}$

(v) $$ \begin{aligned} \alpha^3-\beta^3 &=(\alpha-\beta)^3+3 \alpha \beta(\alpha-\beta) \\ &=(\alpha-\beta)\left\{(\alpha+\beta)^2-\alpha \beta\right\} \end{aligned} $$

(vi) $\alpha^4+\beta^4=\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}^2-2 \alpha^2 \beta^2$

(vii) $\alpha^4-\beta^4=\left(\alpha^2-\beta^2\right)\left(\alpha^2+\beta^2\right)$

$$ =(\alpha-\beta)(\alpha+\beta)\left((\alpha+\beta)^2-2 \alpha \beta\right) $$

(viii) $\alpha^2+\alpha \beta+\beta^2=(\alpha+\beta)^2-\alpha \beta$

(ix) $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}$

(x) $\alpha^2 \beta+\beta^2 \alpha=\alpha \beta(\alpha+\beta)$

(xi) $\left(\frac{\alpha}{\beta}\right)^2+\left(\frac{\beta}{\alpha}\right)^2=\frac{\alpha^4+\beta^4}{\alpha^2 \beta^2}=\frac{\left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2}{\alpha^2 \beta^2}$

$\begin{array}{rl}P(x)=& a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_{1}x+a_0.......(1)\\ & \\ =& a_n\left(x-{\alpha }_1\right)\left(x-{\alpha }_2\right)\dots \left(x-{\alpha }_n\right)\\ & \\ =& a_n[x^n-x^{n-1}\left(\Sigma {\alpha }_1\right)+x^{n-2}\left(\Sigma {\alpha }_1{\alpha }_2\right)+\dots +\\ & (-1{)}^n{\alpha }_1{\alpha }_2{\alpha }_3\dots {\alpha }_n].......(2)\end{array}$

Now comparing the coefficients from above identity (1) and (2) then

Sum of the roots taking one at a time

$=\Sigma {\alpha }_1={\alpha }_1+{\alpha }_2+{\alpha }_3+\dots +{\alpha }_{n-1}+{\alpha }_n=-\frac{a_{n-1}}{a_n}$

Sum of product of roots taken two at a time $=$

$\Sigma {\alpha }_1{\alpha }_2={\alpha }_1{\alpha }_2+{\alpha }_2{\alpha }_3+{\alpha }_3{\alpha }_4+\dots +{\alpha }_{n-1}{\alpha }_n+{\alpha }_n{\alpha }_1$ $=(-1{)}^2\frac{a_{n-2}}{a_n}=+\frac{a_{n-2}}{a_n}$

Sum of product of roots taken three at a time $=$

$\begin{array}{rl}& \Sigma {\alpha }_1{\alpha }_2{\alpha }_3={\alpha }_1{\alpha }_2{\alpha }_3+{\alpha }_2{\alpha }_3{\alpha }_4+{\alpha }_3{\alpha }_4{\alpha }_5+\dots +\\ & \\ & {\alpha }_{n-2}{\alpha }_{n-1}{\alpha }_n=(-1{)}^3\frac{a_{n-3}}{a_n}=-\frac{a_{n-3}}{a_n}\end{array}$

Product of the all roots $=$

${\alpha }_1{\alpha }_2{\alpha }_3\dots {\alpha }_{n-1}{\alpha }_n=(-1{)}^n\frac{a_0}{a_n}$

In quadratic equation $a x^2+b x+c=0$, the term $b^2-4 a c$ is called discriminant of the equation, which plays an important role in finding the nature of the roots. It is denoted by $\Delta$ or $D$.

(a) Suppose $a, b, c \in \mathbf{R}$ and $a \neq 0$ then

(i) If $D > 0 \Rightarrow$ Roots are real and unequal

(ii) If $D=0 \Rightarrow$ Roots are real and equal and each equal to $-b / 2 a$

(iii) If $D < 0 \Rightarrow$ Roots are imaginary and unequal or complex conjugate.

(b) Suppose $a, b, c \in Q$ and $a \neq 0$ then

(i) If $D>0$ and $D$ is perfect square

$\Rightarrow$ Roots are unequal and Rational

(ii) If $D>0$ and $D$ is not perfect square

$\Rightarrow$ Roots are irrational and unequal

Therefore, we can conclude the following results.

1. If $D_1$ and $D_2$ are discriminant of two quadratic equations and $D_1+D_2 \geq 0$ then atleast one $D_1$ and $D_2 \geq 0$.

$\Rightarrow$ Atleast one of the equations has real roots

2. $D_1+D_2<0$

$\Rightarrow$ At least one of $D_1$ and $D_2<0$.

$\Rightarrow$ At least one of the equations has imaginary root.

3. If $D_1$ is the discriminant of the equation $a_1 x^2+b_1 x+c_1=0$ and $D_2$ is the discriminant of the equation $a_2 x^2+b_2 x + c_2=0$ and

$\left(a_1 x^2+\right.$ $\left.b_1 x+c_1\right)\left(a_2 x^2+b_2 x + c_2\right)=0$ .........(1)

(i) $D_1 D_2<0 \Rightarrow D_1>0$ and $D_2<0$ or $D_1<0$ and $D_2>0$

Then equation (1) has two real roots.

(ii) $D_1 D_2>0$

Case I: $D_1>0$ and $D_2>0$

Then equation (1) has four real roots.

Case II: $D_1<0$ and $D_2<0$

Then equation (1) has no real roots.

(iii) $D_1 D_2=0$

Case I: $D_1>0$ and $D_2=0$ or $D_1=0$ and $D_2>0$

Then equation (1) has two equal real roots and two distinct roots.

Case II: $D_1<0$ and $D_2=0$ or $D_1=0$ and $D_2<0$

Then equation (1) has two equal real roots.

Case III: $D_1=0$ and $D_2=0$

Then equation (1) has real roots each repeated twice.