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QUADRATIC EQUATION & EXPRESSIONS
Quadratic Equation
An equation of the form ax2 + bx + c = 0, where a ≠ 0 and a, b, c are real numbers, is called a quadratic
equation. The numbers a, b, c are called the coefficients of the quadratic equation. A root of the quadratic
equation is a number α (real or complex) such that aα2 + bα + c = 0.
The roots of the quadratic equation are given by x =
−b ± b2 − 4ac
.
2a
Basic Results:
‰
‰
‰
‰
The quantity D (D=b2 – 4ac) is known as the discriminant of the quadratic equation.
The quadratic equation has real and equal roots if and only if D = 0 i.e. b2 – 4ac = 0.
The quadratic equation has real and distinct roots if and only if D > 0 i.e. b2 – 4ac > 0.
The quadratic equation has complex roots with non–zero imaginary parts if and only if D < 0 i.e.
b2 – 4ac < 0.
‰ Let α and β be two roots of the given quadratic equation. Then α + β = – b/a and αβ = c/a.
‰ If α and β are the roots of an equation, then the equation is x2 − (α + β) x + αβ = 0.
‰ In a quadratic equation ax2 + bx + c = 0 if a + b + c = 0, then x = 1 is one of the root of the
equation.
‰ If the quadratic equation ax2 + bx + c = 0 is satisfied by more than two numbers (real or
imaginary) then it becomes an identity i.e., a = b = c = 0
x +1 x − 2
+
= 3 (x ≠ 1, - 2).
Illustration 1: Solve
x −1 x + 2
Solution:
x +1 x − 2
+
=3
x −1 x + 2
x +1 x − 2
+
−3 = 0
⇒
x −1 x + 2
( x + 1)( x + 2 ) + ( x − 1)( x − 2 ) − 3 ( x − 1)( x + 2)
=0
⇒
( x − 1)( x + 2 )
⇒ (x + 1) (x + 2) + (x – 1) (x – 2) – 3(x – 1) (x + 2) = 0
⇒ x2 + 3x + 2 + x2 – 3x + 2 – 3(x2 + x – 2) = 0
⇒ − x2 – 3x + 10 = 0
⇒ − (x2 + 3x – 10) = 0
⇒ x2 + 3x – 10 = 0
⇒ (x + 5) (x – 2) = 0 ⇒ x = − 5, or x = 2.
∴ x = −5, 2 are the solutions of the given equation.
Illustration 2: Find if the roots of the equation (a + b − 2c) x2 + (a + c − 2b) x + (b + c − 2a) = 0 are real
or complex, where a, b, c ∈ R.
Solution:
Since (a + b − 2c) + (a + c − 2b) + (b + c − 2a) = 0
⇒ x = 1 is one of the root.
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Hence the roots are real.
Illustration 3:. The sum of squares of two consecutive positive integers is 221. Find the integers.
Solution:
Let x be one of the positive integers. Then the other is x + 1
∴ sum of squares of the integers = x2 + (x + 1)2 = 221
∴ x2 + x2 + 2x + 1 – 221 = 0
2x2 + 2x – 220 = 0
2(x2 + x – 110) = 0
x2 + x – 110 = 0
∴ (x – 10) (x + 11) = 0 ∴ x = 10 or x = −11.
∴ consecutive positive integers are 10 and 11.
Illustration 4:. One side of a rectangle exceeds its other side by 2 cm. If its area is 195 cm2, determine
the sides of the rectangle.
Solution:
Let one side be x cm
Then other side will be (x + 2) cm
Area of rectangle = x (x + 2)
∴ x (x + 2) = 195
x2 + 2x – 195 = 0
x2 + 15x – 13x – 195 = 0
x(x + 15) – 13(x + 15) = 0
(x – 13) (x + 15) = 0
∴ x = 13 or x = −15.
Since, side of the rectangle cannot be negative
∴ x = 13.
∴ sides of rectangle are x, x + 2 = 13 cm, 15 cm.
Illustration 5: The hypotenuse of a right angled triangle is 25 cm. The difference between the lengths of
the other two sides of the triangle is 17 cm. Find the lengths of these sides.
Solution:
Let the length of the shorter side be x cm.
Then, the length of the longer side = (x + 17) cm
∴ AB = x, BC = x + 17, CA = 25
By Pythagoras theorem
AB2 + BC2 = 252 ⇒ x2 + (x + 17)2 = 252
2x2 + 34x – 336 = 0
⇒ x2 + 17x – 168 = 0
(x – 7) (x + 24) = 0 ⇒ x = 7, x = −24
But side of triangle cannot be negative
A
x
B
25
x+17
C
∴x=7
∴ Length of the shorter side x = 7 cm
Let of the longer side = x + 17 = 7 + 17 = 24 cm.
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Illustration 6: Solve the following equation by factorization method
Solution:
4
5
−3 =
x
2x + 3
x ≠ 0, −
3
.
2
4
5
−3 =
x
2x + 3
⇒ (4 − 3x) (2x + 3) = 5x
⇒ 6x2 + 6x − 12 = 0
⇒ x2 + x − 2 = 0
⇒ (x + 2) (x − 1) = 0
⇒ x = −2 or x = 1.
Illustration 7: Find k for which the equation (k − 12)x2 + 2(k − 12) x + 2 = 0 has equal roots.
Solution:
For the given equation
a = k − 12, b = 2(k − 12) c = 2
D = b2 − 4ac
= 4(k − 12)2 − 4(k − 12) (2)
= 4(k −12) (k − 14)
A quadratic equation has equal roots if D = 0
⇒ 4(k − 12) (k − 14) = 0
⇒ k = 12 or k = 14
⇒ k =14 only.
(
)
Illustration 8: If the equation x 2 − ( 2 + m ) x + m2 − 4m + 4 = 0 has coincident roots, then find the
possible values of m.
Solution:
∵ roots are coincident,
discriminant = 0
(
)
∴ ( 2 + m ) = 4 m2 − 4m + 4 .
2
This will give m =
2
or 6 .
3
Illustration 9: Solve 3x + 3-x – 2 = 0.
Solution:
3x + 3−x – 2 = 0
1
⇒ 3x + x − 2 = 0
3
1
⇒t+ −2=0
t
2
⇒ t – 2t + 1 = 0
⇒ (t – 1)2 = 0 ⇒ t = 1
∴ 3x = 1 ⇒ 3x = 30 ⇒ x = 0.
let t = 3x
Illustration 10: If the equation (λ2 − 5λ + 6 )x2 + (λ2 − 3λ + 2)x + (λ2 − 4) = 0 is satisfied by more than two
values of x, then find the parameter λ.
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Solution:
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If an equation of degree two is satisfied by more than two values of x, then it must be an
identity.
λ2 − 5λ + 6 = 0, λ2 − 3λ + 2 = 0, λ2 − 4 = 0
⇒ λ = 2, 3 and λ = 1, 2 and λ = 2, − 2.
Common value of λ which satisfies each condition is λ = 2.
Illustration 11: Find the value of m for which the equation (1 + m)x2 –2(1 + 3m)x + (1 + 8m) = 0 has
equal roots.
Solution:
Given equation is (1 + m)x2 –2(1 + 3m)x + (1 + 8m) = 0.
If roots are equal, then discriminant = 4(1 + 3m)2 –4(1 + m) (1 + 8m) = 0
⇒ m2 – 3m = 0 or m = 0, 3.
Problems based on sum and product of roots
Illustration 12: Find the value of m if the product of the roots of the equation mx2+ 6x + (2m –1) = 0 is –
1.
Solution:
Product of the roots =
2m − 1
= −1
m
⇒ 2m – 1 = –m
⇒ 3m = 1
1
⇒m= .
3
Illustration 13: If the sum of the roots of the equation qx2 + 2x + 3q = 0 is equal to their product, then find
the value of q.
Solution:
Let α, β be the roots
2
∴α+β= −
q
3q
=3
αβ=
q
Since α + β = α β
2
∴ − =3
q
2
⇒q= −
3
Illustration 14: Determine the equation, sum of whose roots is 1 and sum of their squares is 13.
Solution:
Let α, β be roots of the required quadratic equation
∴ α + β = 1, α2 + β2 = 13
(α+ β)2 = α2 + β2 + 2αβ
1 = 13 + 2αβ
2αβ = -12
∴ αβ = -6.
∴ required quadratic equation is x2 – (α + β)x + αβ = 0
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x2 – x – 6 = 0.
Illustration 15: If the roots of the equation ax2 + bx + c = 0 are in the ratio m : n, prove that
mnb2 = ac(m + n)2
Solution:
α:β=m:n
α m
=
⇒
β n
α+β m+n
=
⇒
(Applying Componendo and Dividendo)
α −β m−n
( α + β ) (m + n)
=
2
2
( α − β ) (m − n)
2
2
(α + β)
(m + n)
=
2
2
( α + β ) − 4αβ (m − n )
2
(m + n)
b2 / a 2
=
b2
c ( m − n )2
−4
2
⇒
⇒
⇒
a2
2
a
(m + n)
b2
=
2
b − 4ac ( m − n )2
2
⇒
⇒ b2(m – n)2 = b2 (m + n)2 – 4ac( m + n)2
⇒ 4ac( m + n)2 = b2 [(m + n)2 – (m – n)2]
⇒ 4ac(m + n)2 = b2 [4mn]
⇒ mnb2 = ac(m + n)2
 α 2 β2 
α β
Illustration 16: If α and β be roots of equation ax2 + bx + c = 0, prove that a 
+
 + b  +  = b.
 β
α
β α

Solution:
Here α + β = – b/a and αβ = c/a.
 α 2 β2 
α β
Now, a 
+
 + b  + 
 β
α 
β α

=
(
) (
a α 3 + β3 + b α 2 + β2
)
αβ
3
2
a ( α + β ) − 3αβ ( α + β )  + b ( α + β ) − 2αβ 




=
αβ
  b 3
 b  2
c  b 
c
a  −  − 3  −   + b  −  − 2 
a  a  
a 
 a 
 a 
=
c/a
bc
= a =b.
c /a
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Illustration 17: If α and β are the roots of the equation 2x2 – 3x – 6 = 0, then find the equation whose
roots are α2 + 2, β2 + 2.
Solution:
Here α + β = 3/2, αβ = –6/2 = –3 so that
S = α2 + β2 + 4 = (α + β)2 – 2αβ + 4 =
49
,
4
P = α2β2 + 2(α2 + β2) + 4 = α2β2 + 4 + 2[(α + β)2 – 2αβ ] =
118
.
4
 49 
 118 
Therefore, the equation is x2 – 
x +
=0
4


 4 
⇒ 4x2 – 49x + 118 = 0.
Illustration 18: If the roots of the equation x2 – px + q = 0 differ by unity then find the relation between p
and q.
Solution:
Suppose the equation x2 – px + q = 0 has the roots α + 1
and α then α + 1+ α = p ⇒ 2α = p – 1 . . . . (1)
and (α+1) α = q ⇒ α2 + α = q.
. . . .. (2)
Putting the value of α from (1) in (2), we get
(p − 1)2
p −1
+
=q
4
2
⇒ (p – 1)2 + 2(p – 1) = 4q
⇒ p2 – 1 = 4q ⇒ p2 = 4q + 1.
Alternative: Let α and β be the roots. |α – β| = 1
⇒ (α + β)2 – 4αβ = 1
⇒ p2 – 4q = 1, or p2 = 1+ 4q.
Illustration 19: If p and q are the roots of the equation x2 + px + q = 0, then find the value of p and q.
Solution:
Since p and q are roots of the equation x2 + px + q = 0,
p + q = – p and pq = q
⇒ pq = q
⇒ q = 0 or p = 1.
If q = 0, then p = 0 and if p =1, then q = –2.
Illustration 20: (α, β); (β, γ) and (γ, α) are respectively the roots of x2 − 2px + 2 = 0, x2 − 2qx + 3 = 0 and
x2 − 2rx + 6 = 0. If α, β and γ are all positive, then find the value of p + q + r.
Solution:
We have, αβ = 2, βγ = 3 and γα = 6,
then αβ ⋅ βγ ⋅γα = 2 ⋅ 3 ⋅ 6 ⇒ αβγ = 6 and α, β, γ are all positive.
Thus we have α = 2, β = 1, γ = 3
Also, 2p = α + β = 3, 2q = β + γ = 4 and 2r = γ + α = 5.
Hence p + q + r = 6.
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ASSIGNMENTS
1.
Show that if the roots of the equation (a2 + b2)x2 + 2x (ac + bd) + c2 + d2 = 0 are real, they will be
equal.
2.
Prove that the roots of the quadratic equation ax2 – 3bx – 4a = 0 are real and distinct for all real a
and b, where a ≠ b.
3.
Prove that the roots of the equation bx2 + (b –c)x + (b –c –a) = 0 are real if those
of ax2 + 2bx + b = 0 are imaginary.
4.
If α, β are the roots of the equation x2 – px + q = 0 and α1, β1 be the roots of the equation x2 – qx
1
1
1
1
+
.
+ p = 0, then find
+
+
α1β αβ1 αα1 ββ1
5.
If b ∈ R+, then show that equation ( 2 + b ) x 2 + ( 3 + b ) x + ( 4 + b ) = 0 has no real roots.
6.
Find the real roots of 32x
7.
Let x2 – (m – 3)x + m = 0 (m ∈ R) be a quadratic equation. Find the value of m for which the roots
(i)
are equal,
(ii)
are opposite in sign,
8.
If the roots of the equation (x – a) (x – b) – k = 0 are c and d, then prove that the roots of
(x – c) (x – d) + k = 0 are a and b.
9.
The coefficient of x in the quadratic equation x2 + px + q = 0 was taken as 17 in place of 13. Its
roots were found to be –2, and –15. Find the roots of the original equation.
10.
If α, β, γ are three distinct roots of the equation (3 − a)x2 + (2 + b)x + (c − 1) = 0, then find the
value of (3 − a)52 + (2 + b) 5 + (c − 1).
11.
If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 be equal, then prove that a, b, c are in
arithmetic progression.
12.
Find the number of real roots of the equation (x − 1)2 + (x − 2)2 + (x − 3)2 = 0.
13.
If sum of roots of the equation x2 + (a + 3) x + b = 0, a, b ∈ R is − 2, then find the value of a.
14.
If one root of the equation x2 – x – k = 0 be square of the other, then find the value of k.
15.
If the sum of the roots of the equation (a + 1)x2 + (2a + 3)x + (3a + 4) = 0 is −1, find the product of
the roots.
16.
If sinα, cosα are the roots of equation cx2 + bx + a = 0, then show that b2 − 2ac − c2 = 0.
17.
If α, β be the roots of the equation 3x2 – 6x + 4 = 0, then find the value of
 α 2 β2   α β 
 1 1
+

 +  +  + 2  +  + 3αβ .
β
α
β
α

α β

 
2
−7x + 7
=9.
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18.
Let a, b, c be three distinct positive real numbers, then find the number of real roots of
ax4 + 2bx2 + c = 0.
19.
If x = 2 is the root of the equation ax2 + bx + 2c = 0, then find the value of 2a + b + c.
20.
Solve
21.
Find the solutions set of x4 – 5x2 + 6 = 0.
22.
If α, β are the roots of the equation x2 + px + q = 0, then find out the quadratic equation whose
α
β
roots are 1 + , 1 + .
β
α
23.
If x =
24.
If one root is the square of the other root of equation x2 + px + q = 0, then find the relation
between p and q.
25.
If the quadratic equation ax2 + bx + c = 0 has the roots 0 and 1, then find the value of a + b.
2x
1
3x + 9
+
+
=0.
x − 3 2x + 3 ( x − 3 )( 2x + 3 )
2
and x = −3 are the roots of the equation ax2 + 7x + b = 0, find the values of a and b.
3
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ANSWERS TO ASSIGNMENTS
4.
p2q2x2 – p2q2x + p3 + q3 – 4pq = 0
6.
5
,1
2
7.
(i)
9.
–3, –10
10.
12.
13.
14.
0
zero
−1
(2 ± √5)
15.
16.
17.
18.
19.
20.
2
b2 – 2ac –c2 = 0
8
0
0
x = −1
21.
x = ± 2, ± 3
22.
23.
24.
25.
m ∈ {1, 9}
{
2
2
(ii)
m<0
}
2
qx − p x + p = 0
3, −6
p3 − (3p − 1) q + q2 = 0
0
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