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CBSE Assignments class 09 Mathematics

  Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX 

CBSE Class 10 Maths Formulas Chapter-04 | Quadratic Equations


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
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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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