Quadratic Equation

Results
1. An equation of degree n has n roots, real or imaginary
2. The equation ${\mathrm{ax}}^2$ + bx + C= 0 where a$\ne$0 is called a quadratic equation and its roots are $\frac{-b\pm \sqrt{b^2-4\mathrm{ac}}}{2a}$
3. For a quadratic equation ${\mathrm{ax}}^2$ + bx + c= 0, the sum of roots $\alpha$ + $\beta$= $\frac{-b}{a}$, product of roots $\alpha \beta$= $\frac{c}{a}$
4. Difference of roots of ${\mathrm{ax}}^2$ + bx + c= 0 is |$\frac{\sqrt{b^2-4\mathrm{ac}}}{a}$|
5. For a cubic equation ${\mathrm{ax}}^3$ + ${\mathrm{bx}}^2$ + c= 0, the sum of roots $\alpha$+$\beta$+$\gamma$= $\frac{-b}{a}$, product of roots $\alpha \beta \gamma$= $\frac{-d}{a}$
6. In ${\mathrm{ax}}^2$ + bx + c= 0

(a) If c= 0 one root is zero and other root is -b/a

(b) If b= 0 roots are equal in magnitude and opposite in sign

(c) If a= c roots are reciprocals to each other

(d) If a,b,c are of same sign then both roots are negative

(e) If a&c are of same sign & b is of opposite sign both roots are +ve.

(f) If a&c are of different sign roots are of different signs.

7. The Discriminant $\mathrm{\Delta }$ = $b^2$ - 4ac and if

$\mathrm{\Delta }$ $<$0, the roots are imaginary

$\mathrm{\Delta }$ $\ge$0, roots are real

$\mathrm{\Delta }$ = 0 roots are real and equal

$\mathrm{\Delta }$ = a perfect square, roots are rational

$\mathrm{\Delta }$ $>$0 and $\ne$ a perfect square, roots are irrational

8. ${\alpha }^2$ + ${\beta }^2$ = ${(\alpha +\beta )}^2$ - 2 $\alpha \beta$
9. ${\alpha }^3$ + ${\beta }^3$ = ${(\alpha +\beta )}^3$ - 3 $\alpha \beta$ ($\alpha +\beta )$
10. ${\alpha }^4$ + ${\beta }^4$ = ${({\alpha }^2+{\beta }^2)}^2$ - 2 ${\alpha }^2{\beta }^2$

Nature of roots
$b^2$ - 4ac is called 'discriminant' of ${\mathrm{ax}}^2$ + bx + c= 0 and is denoted by $\mathrm{\Delta }$ or D

(i) If D $>$0 and a perfect square, then the the roots are real, rational and distinct.

(ii) If D $>$ 0 and not a perfect square, then the roots are real, irrational and distinct.

(iii) If D = 0 the roots are real and equal.

(iv) If D $<$ 0, then the roots are complex conjugates.

Factor of ${\mathrm{ax}}^2$ + bx + c= 0

If '$\alpha$' is a root of ${\mathrm{ax}}^2$ + bx + c= 0, then (x-$\alpha$) is a factor of ${\mathrm{ax}}^2$ + bx + c.

The irrational and imaginary roots of a quadratic equation are occure only in conjugate

pairs, if the co-efficients
are real.

Common Roots

Let the quadratic equations be: ${\mathrm{ax}}^2$ + bx + c= 0 and $a^1$ $x^2$+ $b^1$x + $c^1$= 0

(i) If one root is common, then ${({\mathrm{ca}}^1-c^{1}a)}^2$= (${\mathrm{ab}}^1$- $a^1$b) (${\mathrm{bc}}^1$- $b^1$c)

(ii) If both roots are common, then $\frac{a}{a^1}$= $\frac{b}{b^1}$= $\frac{c}{c^1}$

Conditions for the roots to lie in the given interval:

(i) The roots $\alpha$ and $\beta$ lie between $k_1$ and $k_2$ if (a) $\mathrm{\Delta }$ $>$0 (b) f($k_1$)$>$0, f($k_2$)$>$0 (c) $k_1$ $<$ $\frac{-b}{2a}$$<$$k_2$

(ii) Exactly one root of f(x) = 0 lies in the interval ($k_1$, $k_2$) if (a) $\mathrm{\Delta }$ $>$ 0 (b) f($k_1$) f($k_2$) $<$ 0

(iii) The number 'k' lies between the roots of a quadratic equation f(x)= 0 if (a) $\mathrm{\Delta }$ $>$ 0 (b) f(k) $<$ 0

Sign of a Quadratic expression for x$\in$R

(i) ${\mathrm{ax}}^2$+bx+c$>$0 for $\forall$ x$\in$R if a $>$ 0 & $\mathrm{\Delta }$ $<$ 0

(ii) ${\mathrm{ax}}^2$+bx+c$<$0 for $\forall$ x$\in$R if a $<$ 0 & $\mathrm{\Delta }$$>$ 0

Transformation of Equations

(i) To obtain an equation whose roots are the reciprocals of the roots of the given equation is

obtained by replacing 'x' by $\frac{1}{x}$ in the given equation.

(ii) To obtain an equation whose roots are negative of the roots of the given equation, is obtained by

replacing 'x' by '-x' in the given equation.

(iii) To obtain an equation whose roots are squares of the roots of a given equation is obtained by replacing, 'x' by '$\sqrt{x}$' in the given equation.

(iv) To obtain an equation whose roots are cube of the roots of a given equation is obtained by replacing 'x' by '$x^{1/3}$' in th givern equation.

Cubic Equation
If $\alpha$,$\beta$,$\gamma$ are roots of equation ${\mathrm{ax}}^3$+${\mathrm{bx}}^2$+cx+d= 0, a$\ne$0, then $S_1$= $\alpha +\beta +\gamma$= $\frac{-b}{a}$, $S_2$= $\alpha \beta +\beta \gamma +\gamma \alpha$= $\frac{c}{a}$ and $S_3$= $\alpha \beta \gamma$= $\frac{-d}{a}$
The equation can be written as $x^3$- $s_1$ $x^2$+ $s_2$x-$s_3$= 0

Biquadratic Equation
If $\alpha$, $\beta$, $\gamma$, $\delta$ are roots of equation ${\mathrm{ax}}^4$+ ${\mathrm{bx}}^3$+${\mathrm{cx}}^2$+dx+e, a$\ne$0, then $S_1$= $\alpha$+$\beta$+$\gamma$+$\delta$= -b/a, $S_2$= $\alpha \beta$+$\alpha \gamma$+$\alpha \delta$+$\beta \gamma$+$\beta \delta$+$\gamma \delta$= c/a, $S_3$= $\alpha \beta \gamma$+$\beta \gamma \delta$+$\gamma \delta \alpha$+$\alpha \beta \delta$= $\frac{-d}{a}$ and $S_4$= $\alpha \beta \gamma \delta$= $\frac{e}{a}$
The equation can be written as $x^4$- $s_1$ $x^3$+$s_2$ $x^2$-$s_3$x + $s_4$= 0

Inequations
Quadratic inequation: Let f(x) = ${\mathrm{ax}}^2$+bx+c then f(x) $\ge$ 0, f(x) $\le$ 0, f(x) $<$ 0, f(x) $>$ 0 are quadratic inequations.
Method to find its solution:

(i) Obtain $\mathrm{\Delta }$= $b^2$- 4ac of Quadratic inequation

(ii) find sign of $\mathrm{\Delta }$

(iii) If $\mathrm{\Delta }$ $<$ 0, then find sign of 'a'

(a) if a $<$0, then f(x) $<$ 0, $\forall$ $x_{\in }$R

(b) if a $>$0, then f(x) $>$ 0, $\forall$ $x_{\in }$R

(iv) If $\mathrm{\Delta }$ = 0, then find sign of 'a'

(a) if a$<$0, then f(x) $<$0, $\forall$ $x_{\in }$R

(b) if a$>$0, then f(x) $\ge$0, $\forall$$x_{\in }$R

(v) if $\mathrm{\Delta }$ $>$ 0, then

(a) Make coefficient of $x^2$+ve. If it is -ve, multiply both sides of inequation by -1 to make it

positive.

(b) Factorise the expression and write the left hand of the inequation in the form (x-$\alpha$) (x-$\beta$). If (x-$\alpha$)

(x-$\beta$) $<$0, then x lies between '$\alpha$' and '$\beta$'. If (x-$\alpha$) (x-$\beta$) $>$0, then x does not lie between $\alpha$ and $\beta \mathrm{.}$


Rational Algebraic Inequations

The inequations of the form $\frac{P(x)}{Q(x)}$ $\ge$ 0, $\frac{P(x)}{Q(x)}$$\le$ 0, $\frac{P(x)}{Q(x)}$$>$ 0, $\frac{P(x)}{Q(x)}$$<$ 0 where P(x) & Q(x) are

Polynominals in x are called 'Rational Algebraic Inequations'.

Method to find its solution:

(i) Make linear factors of P(x) & Q(x)

(ii) Make coefficient of x positive in each linear factor.

(iii) Obtain critical points by equating each linear factor to zero.

(iv) If there are 'n' critical points, then divide the real line inti (n+1) regions.

(v) Plot these points on a numberline obtained above

(vi) In the right most region the expression $\frac{P(x)}{Q(x)}$ bears positive sign and in other regions the

expression bears alternate +ve and -ve signs, then these critical points form solution.

Some useful Results

1. If the roots of ${\mathrm{ax}}^2$+bx+c= 0 are reciprocal to each other, then c=0

2. If the roots of ${\mathrm{ax}}^2$+bx+c= 0 are equal in magnitude but opposite in sign, then b=0

3. If a & c are of opposite sign, the roots must be of opposite sign.

4. If the roots of ${\mathrm{ax}}^2$+bx+c= 0 are in the ratio m:n, then the condition is mn$b^2$= ac ${(m+n)}^2$

5. If the equation ${\mathrm{ax}}^2$+bx+c= 0 (a$\ne$0) where a,b,c $\in$ R (a) If a+b+c= 0, then its roots are 1, $\frac{c}{a}$ (b) If

a-b+c= 0, then its roots are -1, $\frac{-c}{a}$.

6. (i) If a$>$0, then minimum valve of ${\mathrm{ax}}^2$+bx+c= 0 is $\frac{4\mathrm{ac}-b^2}{4a}$

(ii) If a$<$0, then max. valve of ${\mathrm{ax}}^2$+bx+c is $\frac{4\mathrm{ac}-b^2}{4a}$

7. If $a_1$, $a_2$, $a_3$, .............$a_n$ are all positive, then least valve of ($a_1$+$a_2$+...............+$a_n$) $[\frac{1}{a_1}+\frac{1}{a_2}+\mathrm{........}+\frac{1}{a_n}]$= $n^2$