### Dictionary Rank of a Word | Permutations & Combinations

PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

### Quadratic Equations Class 10 Chapter 4

Quadratic equation Chapter 4 Class 10

Basic concepts on Quadratic Equation class 10, chapter 4  Relationship between the Coefficients and Roots,  Nature of roots, Discriminant, Quadratic Formula, method of completing the square. Complete explanation of quadratic equations

QUADRATIC EQUATION :-
An equation whose degree is 2 is called a quadratic equation.
$General\; Quadratic\; Equation\; is:\; \; ax^{2}+bx+c=0$

Difference between the quadratic equations and quadratic polynomials.

Quadratic equations are very similar to the quadratic polynomials. But they are different from each other because of the following reasons.
 Quadratic Equations Quadratic Polynomials General Quadratic Equations is    ax2 + bx + c = 0 General Quadratic Polynomial is  P(x) = ax2 + bx + c Solutions of quadratic equations are called its roots. Solutions of quadratic polynomials are called its zeroes.

Methods of Solving Quadratic Equations
There are mainly three methods of solving Quadratic equations
1) Factor Method
2) By Quadratic Formula
3) My the method of Completing the Square.

ROOTS:-
Solutions of the quadratic equations are called its roots. A quadratic equation have two roots.

RELATION BETWEEN ROOTS AND COEFFICIENTS:-
$Sum \; of\; the \; roots(S)=\frac{-b}{a}$
$Product \; of\; the \; roots(P)=\frac{c}{a}$
Quadratic Equation From The Roots:-
$x^{2}-(Sum\; of\; the\; roots)\; x+Product\; of\; the \; roots\; =\; 0$
$x^{2}-Sx+P=0$

Nature Of The Roots Of The Quadratic Equation:-
General quadratic equation is
ax2 + bx + c = 0

Discriminant of quadratic equation is denoted by D and is given by
D = b2 - 4ac
$If D> 0 \; then \; roots\; are \; real\; and\; unequal\; or\; distinct\; or\; different)$
$If D= 0 \; then \; roots\; are \; real\; and\; equal$
$If D< 0 \; then \; roots\; are \; not\; real$
$If D\geq 0 \; then \; roots\; are \; \; real$

Quadratic Formula For Solving The Quadratic Equations:-
$x=\frac{-b\pm \sqrt{D}}{2a}\; \; \; or\; \; \; x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
Method of completing the square:-
Write the given quadratic equation
$ax^{2}+bx+c=0$
$Make\; the\; coefficient \; of \; x^{2}\; unity$
$x^{2}+\frac{b}{a}x+\frac{c}{a}=0$
Bring the constant term to the right hand side.
$x^{2}+\frac{b}{a}x=-\; \frac{c}{a}$
$Adding \; \left ( \frac{1}{2}\; of \; the\; coefficient \; of\; x\; \right )^{2}\; on\; both \; side.$
$x^{2}+\frac{b}{a}x+\left ( \frac{b}{2a} \right )^{2}=\frac{b^{2}}{4a^{2}}-\; \frac{c}{a}$
Completing the square
$\left ( x+\frac{b}{2a} \right )^{2}=\left ( \frac{b^{2}-4ac}{4a^{2}} \right )$
Taking the square root on both side
$\left ( x+\frac{b}{2a} \right )=\pm \frac{\sqrt{b^{2}-4ac}}{2a}$
Find the value of x
$x=-\frac{b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$x=\frac{-b\pm \sqrt{D}}{2a}$
$x=\frac{-b+\sqrt{D}}{2a}\: \: or\: \: \frac{-b-\sqrt{D}}{2a}$
Example
 Find the roots of  5x2 - 6x - 2 = 0 by the method of completing the square Given equation is   5x2 - 6x - 2 = 0 D = b2 - 4ac  ⇒ D = (-62 ) - 4 x 5 x -2 = 36+40 = 76 > 0 ⇒ Roots are real and distinct ⇒ Roots of Q. E. are exists Now we apply method of completing the square as follows Step (1) Make the coefficient of x2 unity ( divide the equation by 5) $x^{2}-\frac{6}{5}x-\frac{2}{5}=0$ Step (2) Bring the constant term to the RHS $x^{2}-\frac{6}{5}x=\frac{2}{5}$ $Step\: (3)\: \:\: \:\: \:\: \:\: \:\: \:\: \:\: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .$$Adding\: \left [ \frac{1}{2}\left ( Coefficient\: of\: x \right ) \right ]^{2}\: on\: both\: side$  $or\: \: Adding\: \left [ \frac{1}{2}\times \frac{6}{5} \right ]^{2} = \left ( \frac{3}{5} \right )^{2} on\: both\: side$   $x^{2}-\frac{6}{5}x+\left ( \frac{3}{5} \right )^{2}=\frac{2}{5}+\left ( \frac{3}{5} \right )^{2}$  Step (4)  Completing the square and find the square root $\left ( x-\frac{3}{5} \right )^{2}=\frac{2}{5}+\frac{9}{25}$$=\frac{2\times 5+9\times 1}{25}=\frac{19}{25}$$x-\frac{3}{5}=\frac{\pm \sqrt{19}}{5}$$x=\frac{3}{5}\: \frac{\pm \sqrt{19}}{5}$$x=\frac{3\pm \sqrt{19}}{5}$$x=\frac{3+ \sqrt{19}}{5},\: \: \: \frac{3- \sqrt{19}}{5}$

Method of solving the Q. E. by using quadratic formula
 Find the roots of  5x2 - 6x - 2 = 0 by the Quadratic Formula Given equation is   5x2 - 6x - 2 = 0 D = b2 - 4ac  ⇒ D = (-62 ) - 4 x 5 x -2 = 36+40 = 76 > 0 ⇒ Roots are real and distinct ⇒ Roots of Q. E. are exists Now we apply the quadratic formula $x=\frac{-b\pm \sqrt{D}}{2a}=\frac{-(-6)\pm \sqrt{76}}{2\times 5}$ $x=\frac{6\pm \sqrt{4\times 19}}{10}=\frac{6\pm 2\sqrt{19}}{2\times 5}$ $x=\frac{2\left (3\pm \sqrt{19} \right )}{10}=\frac{3\pm \sqrt{19} }{5}$ $x=\frac{3+ \sqrt{19} }{5},\; \; \; \frac{3- \sqrt{19} }{5}$

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