QUADRATIC EQUATIONIdentities

An equation is called identity if and only if all real numbers are the solution of its.

Problem: Find all value of p for (p2-3p+2)x2+ (p2-5p+4)x + p - p2 = 0

is an identity.

Sol: p2-3p+2 = 0 p2-5p+4 = 0 p-p2 = 0 ⇒ (p-2) (p-1) = 0 ⇒ (p-4) (p-1) = 0 ⇒p (1-p)=0 ⇒ P = 2, 1 ⇒ p=4, 1 ⇒ p=0, 1 ∴p=1 ansProblem: Number of solution of equation is

( x−a )(x−b)( c−a )(c−b)

+ ( x−b )(x−c)(a−b )(a−c)

+ ( x−c )(x−a)(b−c )(b−a)

= 1a) 0 b) 2 c) 3 d) ∞Ans: (d)

Sol: max power of x is 2 but it 3 roots can be easily seen x=a, x=b, x=c. i.e. it is not a quadratic equation, it is an identity thus all real numbers are its solution.

QUADRATIC EQUATION

a x2+bx+c=0Where a≠0

is a quadratic equation and the value of x which is satisfied this equation is called its ‘Root’.

NOTE: - It has always two roots, may be real & imaginary; equal & unequal.

Quadratic Expression:

y = a x2+bx+c

Where a≠0 and x, y are variable and a, b, c are any constant

Sridharacharya Method for finding root of quadratic equation

a x2+bx+c=0⇒x=−b±√b2−4ac2a

D = b2−4 ac is called its ‘Discriminant’α = −b2a + √D

2a

β = −b2a - √D2a

where α, β are its rootsα+β=−b

2aα . β= c

a

a x2+bx+c=0Divided by a both side;

x2−(−ba ) x+( ca )=0⇒a x2−(α+ β)x+α .β=0

Problem: If 𝜶 & 𝜷 are roots of the quadratic equation a x2+bx+c=0 then find the value of(i) α 2 + β2

(ii) α 3 + β3

(iii) |𝛼 – 𝛽|(iv) (aα+b )−4 + (aβ+b )−4

Ans:

(i)b2

a2 - 2ca

(ii)−b3

a3 + 3bc

a2

(iii) √D¿a∨¿¿

(iv)α 4

c4 + β4

c4

Problem: If 𝜶 & 𝜷 are roots of the quadratic equation a x2+bx+c=0 then find the quadratic equation which have roots i. 𝛼+2, 𝛽+2ii. 1+α1−α , 1+β

1−βiii. αα+3 , β

β+3iv. 1α

,1β

Sol: (i) Method 1st

Required equation is

a x2−(α+2+β+2 ) x+(α+2 ) .(β+2)=0⇒a x2−(α+β+4 ) x+α .β+2 (α+ β )+4=0

⇒a x2−(−ba ) x+( ca )+2(−ba )+4=0

Method 2nd (using symmetry)Suppose that p & q are the roots of required equation thenp=α+2 ⇒α=p-2q=β+2 ⇒β=q-2∵ α & β are roots of a x2+bx+c=0∴ a α2+bα+c=0 &a β2+bβ+c=0 p⇒a (x−2)2+b(x−2)+c=0 q

Ans: (i) (x−2)2+b(x−2)+c=0(ii) a (x−1)2+bx (x2−1)+c (x+1)=0(iii) 3a x2+3bx (1−x)+c(1−x)=0(iv) c x2+bx+a=0

Nature of Roots

NOTE: - It is fouls statement that if D≥0 then roots are real in all conduction. It is true if and only if D≥0 with a, b & c are real numbers

e.g. ℩x2 - 5x + =0℩

Nature of Roots

D=0 D≠0 Both roots are equal & real Both roots are different If a, b, c ∈ R D>0 D<0 Roots are real Roots are imaginary

NOTE: - In this conduction roots are conjugates If a, b, c ∈ Q (rational)

If D is a square of any rational number If D is not a square of any rational number then roots are also rational then roots are also irrational NOTE:- In this conduction, get irrational roots are conjugates

NOTE:- If a=1, b & c are integers and D is a square of any rational number then both roots of equation

a x2+bx+c=0 are integers.

Problem: If a, b and c are real then prove that equationx2−2ax+a2−b2−c2=0

have real and unequal roots.

Problem: If a, b and c are rational then prove that equation(a+c−b¿ x2+2c x+(b+c−a)=0

have rational roots.

Condition for common roots of two given quadratic equation

a1 x2+b1 x+c1=0

a2 x2+b2 x+c2=0

# Both roots are common if and only if

a1a2

=b1b2

=c1c2

# For one common rootLet α is a common root of given two equation

∴a1α 2+b1α+c1=0 a2α

2+b2α+c2=0From cross multiplication;

α 2

b1 c2−b2 c1=

αc1a2−c2a1

= 1a1b2−a2b1

⇒ α=b1 c2−b2 c1c1a2−c2a1=c1a2−c2a1a1b2−a2b1

NOTE:- If a1 , b1 , c1 and a2 , b2 , c2 are real and both equation has one common root then other root is also common.

Problem: If x2−5 x+6=0 and x2+2kx−3=0 have a common root then find the value of k. Ans: k=1

Problem: