# Quadratic Equations - Class 10 : Notes

Notes for quadratic equations chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

**Â **

**(1) A polynomial of degree 2 is called a quadratic polynomial. The general form of a quadratic polynomial is , where a, b, c are real number such that a 0 and x is a real variable.
**

**, where are real number. So given equation is quadratic polynomial.**

*For Example:Â***(2) If is a quadratic polynomial and is a real number, then is known as the value of the quadratic polynomial
**

**in that equation if then . So is a value of quadratic polynomial**

*For Example:Â***(3) A real number is said to be a zero of quadratic polynomial , if
**

*For Example:Â*If then , So is the zero of polynomial.

**(4) If is a quadratic polynomial, then i.e., , is called a quadratic equation.
**

**is a quadratic polynomial, then i.e., , is called a quadratic equation.**

*For Example:Â***(5) A real number is said to be a root of the quadratic equation .
**

**In other words,Â Â is a root of if and only if is a zero of the polynomial .**

**Suppose quadratic equation is .**

*For Example:Â*If we put then , So is a root of that given equation so here .

**(6) If , is factorizable into a product of two linear factors,Â then the roots of the quadratic equation can be found by equating each factor to zero.
**

**The Given equation is**

*For Example:Â***Now, Solving the above equation using factorization method.**

*Â*

(3x + 1) (3x - 2) =0

Â (3x + 1) = 0 orÂ (3x - 2) = 0

3x = -1 or 3x = 2

Â Â orÂ

Hence,Â and Â Â are the two roots of the given equation

**(7) The roots of a quadratic equation can also be found by using the method of completing the square.**

*The Given equation is -*

**For Example**:Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Dividing through out by 2

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Shifting the constant term to the right hand side.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Adding square of the half of coefficient of x on the both side.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Taking square root of both sides-

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Hence x= 3, and x=1/2 are the two root of the given equation

**(8) The roots of the quadratic equation , can be found by using the quadratic formula , provided that .**

*For Example:*

*Â***the given equation in the form ofÂ**

*Â***Where a=Â âˆš3, b=10 c= 8âˆš3**

*,*

Therefore, the discriminant-Â

D= (10)

^{2}â€“ 4 x âˆš3 x (-8âˆš3)

D= 100 + 96

D= 196

Since, D > 0

Therefore, the roots of the given equation are real and distinct.

The real roots Î± and Î² are given by,

;Â

For,Â

Hence Â and Â are the two root of the given equation.

(9) Nature of the roots of quadratic equation , depends upon the value of , which is known as the discriminate of the quadratic equation.

** For Example:Â **Value of can be (i) 0" /> , (ii) (iii) .

**(10) The quadratic equation , has:
**

**(i) Two distinct real roots, if D ba-4ac 0 two equal roots i.e. coincident real roots if 0" />**

**16x**

*For Example: Â*^{2}= 24x + 1

16x

^{2}â€“ 24x â€“ 1 = 0

The given equation is of the form of ax

^{2}+ bx + c = 0, where a = 16, b = -24, c = -1

Therefore, the discriminant-Â D = b

^{2}â€“ 4ac

D= (-24)

^{2}â€“ 4 x 16 x (-1)

D= 576 + 64

D= 640

Since, D > 0

Therefore, the roots of the given equation are real and distinct.

The real roots Î± and Î² are given by,

For,Â

Hence Â and Â are the two root of the given equation.

(ii) Two equal roots i.e. coincident real roots, if .

** For Example:Â **2x

^{2}- 2âˆš6x + 3 = 0

The given equation is of the form of ax

^{2}+ bx + c = 0, where a = 2, b = - 2âˆš6, c = 3

Therefore, the discriminant-Â D = b

^{2}â€“ 4ac

= (- 2âˆš6)

^{2}â€“ 4 x 2 x 3

= 24 - 24

= 0

Since, D = 0

Therefore, the roots of the given equation are real.

The real and equal roots are given by Â Â Â and Â .

â‡’

**(iii) No real roots, if .
**

**The given equation is**

*For Example:Â*x

^{2}+ x + 2 = 0

The given equation is of the form of ax

^{2}+ bx + c = 0, where a = 1, b = 1, c = 2

Therefore, the discriminant

D = b

^{2}â€“ 4ac

D= (1)

^{2}â€“ 4 x 1 x 2

D= 1 - 8

D= -7

Since, D < 0

Therefore, the given equation has not real roots.

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