Surds Must Solve Questions
Question 1.
If let us calculate simplified of i.
and ii.
Answer:
Formula used.
(a + b)2 = a2 + b2 + 2ab
(a + b)3 = a3 + b3 + 3ab(a + b)
Let a = m and b =
(a + b)2 = a2 + b2 + 2ab
(m + )2 = m2 +
2 + 2 Γ m Γ
(β3)2 = m2 + + 2 Γ 1
3 = m2 + + 2
m2 + = 3 β 2 = 1
(a + b)3 = a3 + b3 + 3ab(a + b)
(m + )3 = m3 +
3 + 2 Γ m Γ
Γ (m +
)
(β3)3 = m3 + + 2 Γ 1 Γ (β3)
3β3 = m3 + + 2β3
m3 + = 3β3 β 2β3 = β3[3 β 2]
= β 3
Question 2.
Let us show that,
Answer:




= 2β15
Question 3.
Let us simplify
Answer:

]
]
]
=
=
Question 4.
Let us simplify
Answer:
Simplifying part 1
= β35 β β14
Simplifying part 2
= β35 β β10
Simplifying part 3
= β14 β β10
Putting values we get;
[β35 β β14] β [β35 β β10] + [β14 β β10]
β35 β β14 β β35 + β10 + β14 β β10
= 0
Question 5.
Let us simplify
Answer:

Simplifying 1st part by rationalizing the expression by multiplying and dividing by 2 + β2

= =
= 4β3 + 2β6
Simplifying 2nd part by rationalizing the expression by multiplying and dividing by 4β3 + β18

=
Simplifying 3rd part by rationalizing the expression by multiplying and dividing by 3 + β12

= = –
Putting all values we get;
4β3 + 2β6 β (4β3 β β18) + (3β2 + β24)
4β3 + β6 Γ 22 β (4β3 β β18) + (β2 Γ 32 + β24)
4β3 + β24 β (4β3 β β18) + (β18 + β24)
= 2β24 = 4β6Hence,
= 4β6
Question 6.
Let us simplify
Answer:
Simplifying 1st part

= -β6 + β12
Simplifying 2nd part

= β18-β6
Simplifying 3rd part

= -β12 + β18
Putting all values we get;
(-β6 + β12) β (β18 – β6) + (-β12 + β18)
-β6 + β12 β β18 + β6 β β12 + β18
= 0
Question 7.
If x = 2, y = 3 and z = 6, let us write the calculating the value of
Answer:

Putting value x = 2, y = 3, z = 6;

Simplifying 1st part

= -β6 + β12
Simplifying 2nd part

= β18-β6
Simplifying 3rd part

= -β12 + β18
Putting all values we get;
(-β6 + β12) β (β18 – β6) + (-β12 + β18)
-β6 + β12 β β18 + β6 β β12 + β18
= 0
Question 8.
If let us calculate simplified value of
Answer:
x = β7 + β6
then;
=
By simplifying
=
=
=
=
= β7 β β6
Hence;
x β = (β7 + β6) β (β7 β β6)
= 2β6
Question 9.
If let us calculate simplified value of
Answer:
x = β7 + β6
then;
=
By simplifying
=
=
=
=
= β7 β β6
Hence;
x + = (β7 + β6) + (β7 β β6)
= 2β7
Question 10.
If let us calculate simplified value of
Answer:
Formula used.
(a + b)2 = a2 + b2 + 2ab
x + = 2β7
(a + b)2 = a2 + b2 + 2ab
Put a = x and b =
(x + )2 = x2 +
2 + 2 Γ x Γ
(2β7)2 = x2 + + 2
4 Γ 7 = x2 + + 2
28 = x2 + + 2
x2 + = 28 β 2 = 26
Question 11.
If let us calculate simplified value of
Answer:
Formula used.
(a + b)3 = a3 + b3 + 3ab(a + b)
If
x + = 2β7
(a + b)3 = a3 + b3 + 3ab(a + b)
Put a = x and b =
(x + )3 = x3 +
3 + 3 Γ x Γ
Γ (x +
)
(2β7)3 = x3 + + 3 Γ (2β7)
56β7 = x3 + + 6β7
x3 + = 56β7 β 6β7
x3 + = β7 [56 β 6]
= 50β7
Question 12.
Let us simplify :
If the simplified value is 14, let us write by calculating the value of x.
Answer:





4x2 β 2
If 4x2 β 2 = 14
4x2 = 14 + 2 = 16
x2 = = 4
x = β4 = Β±2
Question 13.
If and
let us calculate the followings :
Answer:
a + b =
a + b = =
=
= 3
a-b =
a-b = =
=
= β5
ab = = 1
=
=
=
Putting values we get;
=
Question 14.
If and
let us calculate the followings :
Answer:
a + b =
a + b = =
=
= 3
a-b =
a-b = =
=
= β5
=
=
Question 15.
If and
let us calculate the followings :
Answer:
a + b =
a + b = =
=
= 3
a-b =
a-b = =
=
= β5
ab = = 1
=
=
=
Putting values we get;
=
Question 16.
If and
let us calculate the followings :
Answer:
a + b =
a + b = =
=
= 3
a-b =
a-b = =
=
= β5
ab = = 1
(a + b)3 = a3 + b3 + 3ab(a + b)
a3 + b3 = (a + b)3 – 3ab(a + b)
= (3)3 β 3 Γ 1 Γ 3
= 27 β 9 = 18
(a-b)3 = a3-b3-3ab(a-b)
a3-b3 = (a-b)3 + 3ab(a-b)
= (β5)3 + 3 Γ 1 Γ (β5)
= 5β5 + 3β5
= β5 [5 + 3]
= 8β5

Question 17.
If let us calculate the simplified value of
Answer:
If x = 2 + β3
Then;

Simplifying it we get;

=
= 2-β3
x β = 2 + β3 β [2 – β3]
= 2β3
Question 18.
If let us calculate the simplified value of
Answer:
If y = 2-β3
Then;

Simplifying it we get;

=
= 2 + β3
y + = 2 β β3 + [2 + β3] = 4
(y + )2 = y2 + [
]2 + 2 Γ y Γ
(4)2 = y2 + []2 + 2
y2 + []2 = 16 β 2 = 14
Question 19.
If let us calculate the simplified value of
Answer:
Formula used.
(a-b)3 = a3-b3-3ab(a-b)
If x = 2 + β3
Then;

Simplifying it we get;

=
= 2-β3
x β = 2 + β3 β [2 – β3]
= 2β3
(x β )3 = x3–
3-3 Γ x Γ
Γ (x β
)
(x β )3 = x3–
-3 Γ 1 Γ (x β
)
(2β3)3 = x3– -3 Γ (2β3)
x3 – = 24β3 + 6β3
x3 – = 30β3
Question 20.
If let us calculate the simplified value of
Answer:
x = 2 + β3
y = 2 β β3
xy = (2 + β3) Γ ( 2-β3)
xy = (2)2 β (β3)2
= 4 β 3
= 1
=
= 1 + 1 = 2
Question 21.
If let us calculate the simplified value of
Answer:
Formula used.
(a β b)2 = a2 β 2ab + b2
3x2 β 5xy + 3y2
Add and subtract xy to the equation.
3x2 β 5xy + 3y2 [ + xy β xy]
3x2 β 6xy + 3y2 + xy
3[x2 β 2xy + y2] + xy
3[x-y]2 + xy
x = 2 + β3
y = 2 β β3
xy = (2 + β3) Γ ( 2-β3)
xy = (2)2 β (β3)2
= 4 β 3
= 1
x β y = 2 + β3 β [2-β3]
x β y = 2β3
Putting the values we get;
3[2β3]2 + 1
3[12] + 1
36 + 1 = 37
Question 22.
If and xy = 1, let us show that
Answer:
Formula used.
(a β b)2 = a2 β 2ab + b2
x =
xy = 1
y =
y = =
x + y = +
=
=
= = 5
x-y = β
=
=
= = β21

Add and Subtract xy both on numerator and denominator
=
Putting values we get;
=
Hence proved.
Question 23.
Let us write which one is greater of and
Answer:
Formula used.
(a β b)2 = a2 β 2ab + b2
1st value is β7 + 1
Its square is
(β7 + 1)2 = (β7)2 + 12 + 2 Γ 1 Γ β7 = 7 + 1 + 2β7 = 8 + 2β7
2nd value is β5 + β3
Its square is
(β5 + β3)2 = (β5)2 + (β3)2 + 2 Γ β3 Γ β5 = 5 + 3 + 2β15 = 8 + 2β15
If 7<15
Then β7 < β15
Then 8 + β7 < 8 + β15
Then β(8 + β7) < β(8 + β15)
β΄ β7 + 1 < β5 + β3
Question 24.
If the value of
is
A. 2
B.
C. 4
D.
Answer:
If x = 2 + β3
Then;

Simplifying it we get;

=
= 2-β3
x + = 2 + β3 + [2 – β3]
= 4
Question 25.
If and
then the value of pq is
A. 2
B. 18
C. 9
D. 8
Answer:
p + q = β13
p = β13 β q
pβq = β5
(β13 β q) β q = β5
2q = β13 – β5
q =
p = β13 β q = β13 β =
pq = Γ
=
=
= 2
Question 26.
If and
the value of (a2 + b2) is
A. 8
B. 4
C. 2
D. 1
Answer:
a + b = β5
a = β5 β b
aβb = β3
(β5 β b) β b = β3
2b = β5 – β3
b =
a = β5 β b = β5 β =
ab = Γ
=
=
(a + b)2 = a2 + b2 + 2ab
(β5)2 = a2 + b2 + 2 Γ
a2 + b2 = 5 β 1 = 4
Question 27.
If we subtract from
the value is
A.
B.
C.
D. None of this
Answer:
β125 = β (5 Γ 5 Γ 5) = 5β5
β 125 β β 5
= 5β5 β β5
= β5 [5-1]
= 4β5
= β((4 Γ 4) Γ 5)
= β80
Question 28.
The product of is
A. 22
B. 44
C. 2
D. 11
Answer:
(5-β3)(β3-1)(5 + β3)(β3 + 1)
(5-β3)(5 + β3)(β3-1)(β3 + 1)
(52 β (β3)2)((β3)2 β 12)
(25 β 3)(3 β 1)
22 Γ 2 = 44
Question 29.
Let us write whether the following statements are true or false :
i. and
are similar surds
ii. is a quadratic surd.
Answer:
(i) True.
β75 = β(5 Γ 5 Γ 3) = 5β3
β147 = β(7 Γ 7 Γ 3) = 7β3
β3 is common on both surds
(ii) False
Ο itself is an irrational number
hence;
Square root of Ο is not a surd.
Question 30.
Let us fill up the blank:
i. a _________ number (rational/ irrational)
ii. Conjugate surd of is ________.
iii. If the product and sum of two quadratic surds is a rational number, then the surds are _________ surds.
Answer:
(a) Irrational
As β11 is irrational number
Multiplying it with 5
Also get irrational number
(b) β3 + 5
Conjugate surds are surds which are having same terms but having different symbol( + to- and β to + ) in between both the terms.
(c) Conjugate Surds
While product of conjugate surds
They get square to become rational number
While sum of conjugate surds
They get cancel to become rational number
Question 31.
If x = 3 + 2β2 let us write the value of .
Answer:
If x = 3 + 2β2
Then;

Simplifying it we get;

=
= 3-2β2
x + = 3 + 2β2 + [3 β 2β2]
= 6
Question 32.
Let us write which one is greater of β15 + β3 and β10 + β8
Answer:
1st value is β15 + β3
Its square is
(β15 + β3)2 = (β15)2 + (β3)2 + 2 Γ β15 Γ β3 = 15 + 3 + 2β45 = 18 + 2β45
2nd value is β10 + β8
Its square is
(β10 + β8)2 = (β10)2 + (β8)2 + 2 Γ β10 Γ β8 = 10 + 8 + 2β80 = 18 + 2β80
If 45<80
Then β45 < β80
Then 18 + β45 < 18 + β80
Then β(18 + β45) < β(18 + β80)
β΄ β15 + β3 < β10 + β8
Question 33.
Let us write two mixed quadratic surds of which product is a rational number.
Answer:
Two mixed surds are
5β6 and 7β6
Multiplying both
5β6 Γ 7β6
35 Γ 6 = 210
Which is a rational number.
Question 34.
Let us write what should be subtracted from β72 to get β32
Answer:
Let the number be x
β72 β x = β32
6β2 β x = 4β2
x = 6β2 β 4β2
= β2 [6 β 4]
= 2β2
Question 35.
Let us write simplified value of
Answer:
Simplifying part 1
= β2-1
Simplifying part 2
= β3-β2
Simplifying part 3
= β4-β3
Adding all we get;
β2-1 + β3-β2 + β4-β3
= β4 β 1
= 2 β 1 = 1









