Page 9 - CHAPTER 4 (Quadratic equations)

CHAPTER 4
QUADRATIC EQUATIONS

√3+1 √3-1 √3+ 1 √3- 1
⇒ (x- ) ²= ( ) ² ⇒x- = ±
2 2 2 2

√3+ 1 √3- 1
⇒ x= ± ⇒x= √3, 1
2 2

Hence, the roots are √3 and 1.

Solution by quadratic formula “Sreedhar Acharya’s Rule”

Consider quadratic equation ax²+bx+c=0, a≠0 then x= -b ± √b²-4ac
2a

∴ The roots of x are

-b+ √b²-4ac -b- √b²-4ac
x= and
2a 2a

⇒ x= -b+ √D or, x= -b- √D , where D = b² – 4ac
2a 2a

Thus, if D = b² – 4ac ≥ 0, then the quadratic equation ax²+bx+c=0 has

real roots given by α= -b+ √D and β= -b- √D
2a 2a

Discriminant: If ax²+bx+c=0, a ≠0 (a, b, c ∈R) is a quadratic equation,

then the expression b² –4ac is known as its discriminant and is

generally denoted by D or ∆.

Example 8: Solve the quadratic equation ² − 6 + 4 = 0 by using

quadratic formula (Sreedharacharya’s Rule).

Solution:

On comparing the given equation ² − 6 + 4 = 0 with the standard

quadratic equation ² + + = 0, we get a = 1, b = –6, c = 4

Hence the required roots are

2
-(-6)±√(-6) -4(1)(4) 6±√36-16 6±√20 6±√4×5
x= = = =
2(1) 2 2 2

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