### Mathematics Lab Manual Class 10

Mathematics Lab Manual Class X   lab activities for class 10 with complete observation Tables strictly according to the CBSE syllabus also very useful & helpful for the students and teachers.

### CBSE Class 10 Maths Formulas Chapter-04 | Quadratic Equations

Quadratic equation Chapter 4 Class 10

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

An equation whose degree is 2 is called a quadratic equation.
General Quadratic Equation is  ax2 + bx + c = 0
Here "a" is the coefficient of x2 ,
"b" is the coefficient of x and
"c" is the constant term.

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.

There are mainly three methods of solving Quadratic equations
1) Factor Method
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:-

$Sum \; of\; the \; roots(S)=\frac{-b}{a}$

$Product \; of\; the \; roots(P)=\frac{c}{a}$

x2 – (Sum of roots)x + Product of roots = 0

x2 – Sx + P = 0

Nature Of The Roots Of The Quadratic Equation:-
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 ≥ 0  then  roots are real

$x=\frac{-b\pm \sqrt{D}}{2a}\; \; \; or\; \; \; x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Method of completing the square:-

ax2 + bx + c = 0
Make the coefficient of x2 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 = (-6)2  - 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 = (-6)2  - 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}$ $x=\frac{3+ \sqrt{19} }{5},\; \; \; \frac{3- \sqrt{19} }{5}$