The DAV Class 8 Maths Solutions and DAV Class 8 Maths Chapter 7 Brain Teasers Solutions of Algebraic Identities offer comprehensive answers to textbook questions.
DAV Class 8 Maths Ch 7 Brain Teasers Solutions
Question 1A.
Tick (✓) the correct option.
(i) The factors of (1 – 6z + 9z2) are
(a) (1 + 3z) (1 – 3z)
(b) (z + 3)2
(c) (3z – 1)2
(d) (z – 3)2
Solution:
(c) (3z – 1)2
1 – 6z + 9z2 = (1)2 – 2 . 1 . 3z + (3z)2
= (1 – 3z)2
= (3z – 1)2
(ii) Which of the following is an algebraic identity?
(a) (a – b)2 = a2 + 2ab – b2
(b) (a – b)2 = a2 – 2ab – b2
(c) (a – b)2 = a2 – b2
(d) (a – b)2 = a2 – 2ab + b2
Solution:
(d) (a – b)2 = a2 – 2ab + b2
L.H.S = (a – b)2
= (a – b) (a – b)
= a (a – b) – b (a – b)
= a2 – ab – ab + b2
= a2 – 2ab + b2
= R.H.S.
(iii) Factorised form of y2 + 3y + y + 3 is
(a) (y + 2) (y + 1)
(b) (y + 3)2
(c) (y + 1) (y + 3)
(d) y2 + 32
Solution:
(c) (y + 1) (y + 3)
y2 + 3y + y + 3 = y (y + 3) + 1 (y + 3)
= (y + 3) (y + 1)
(iv) Factorised form of p2 – 17p – 38 is
(a) (p – 19) (p + 2)
(b) (p – 19) (p – 2)
(c) (p + 19) (p + 2)
(d) (p + 19) (p – 2)
Solution:
(a) (p – 19) (p + 2)
p2 – 17p – 38 = p2 – 19p + 2p – 38
= p (p – 19) + 2 (p – 19)
= (p + 2)(p – 19)
(v) A side of the square with area (4x2 + 12x + 9) square units is
(a) (4x + 3) units
(b) (4x + 9) units
(c) (2x + 3) units
(d) (x + 3) units
Solution:
(c) (2x + 3) units
Area of squre = 4x2 + 12x + 9
= (2x)2 + 2 . 2x . 3 + (3)2
= (2x + 3)2 sq. units
Thus, the side of the square is (2x + 3) units.
Question 1B.
Answer the following questions.
(i) Find the product using suitable identity. (a + 3) (a – 3) (a2 + 9)
(ii) If x2 + \(\frac{1}{x^2}\) = 6, find the value of x4 + \(\frac{1}{x^4}\).
(iii) Find the value of k if: xy2k = (4xy + 3y)2 – (4xy – 3y)2
(iv) If a2 + b2 = 9 and ab = 4.
Find the value of 3 (a + b)2 – 2(a – b)2.
(v) If a + b + c = 12 and a2 + b2 + c2 = 64, find the value of ab + bc + ac.
Solution:
(i) (a + 3) (a – 3) (a2 + 9) = {a2 – (3)2} (a2 + 9)
[Using identity (a + b) (a – b) = a2 – b2]
= (a2 – 9) (a2 + 9)
= (a2)2 – (9)2
= a4 – 81.
(ii) Given:
x2 + \(\frac{1}{x^2}\) = 6
Squaring both sides, we get
(x2 + \(\frac{1}{x^2}\))2 = (6)2
(x2)2 + 2 . x2 . \(\frac{1}{x^2}\) + (\(\frac{1}{x^2}\))2 = 36
x4 + 2 + \(\frac{1}{x^4}\) = 36
⇒ x4 + \(\frac{1}{x^4}\) = 36 – 2
∴ x4 + \(\frac{1}{x^4}\) = 34
(iii) We have xy2k = (4xy + 3y)2 – (4xy – 3y)2
xy2k = (4xy + 3y)2 – (4xy – 3y)2
⇒ xy2k = (4xy + 3y + 4xy – 3y) (4xy + 3y – 4xy + 3y)
⇒ xy2k = 8xy × 6y
⇒ xy2k = 48xy2
⇒ k = 48
(iv) Given a2 + b2 = 9 and ab = 4.
Now, 3(a + b)2 – 2(a – b)2 = 3(a2 + 2ab + b2) – 2(a2 – 2ab + b2)
= 3a2 + 6ab + 3b2 – 2a2 + 4ab – 2b2
= (a2 + b2) + 10 ab
= 9 + 10 × 4
= 9 + 40
= 49
(v) Given: a + b + c = 12
and a2 + b2 + c2 = 64
∵ (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
⇒ (12)2 = 64 + 2(ab + bc + ca)
⇒ 144 = 64 + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = 144 – 64
⇒ ab + bc + ca = \(\frac{80}{2}\)
∴ ab + bc + ca = 40.
Question 2.
Find the following products using a suitable identity:
(i) (7x – 9y) (7x – 9y)
(ii) (6ax + 5by) (6ax – 5by)
(iii) \(\left(\frac{3}{4} p^2+\frac{2}{3} q^2\right)\left(\frac{3}{4} p^2+\frac{2}{3} q^2\right)\)
(iv) (x + \(\frac{3}{4}\)y – 4z)2
(v) (x + 2y) (x – 2y) (x2 + 4y2)
(vi) \(\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)\left(x^4+\frac{1}{x^4}\right)\)
Solution:
(i) (7x – 9y) (7x – 9y) = (7x – 9y)2
= (7x)2 – 2 (7x) (9y) + (9y)2
[Using (a – b)2 = a2 – 2ab + b2]
= 49x2 – 126xy + 81y2
(ii) (6ax + 5by) (6ax – 5by)
= (6ax)2 – (5by)2
[Using (a + b) (a – b ) = a2 – b2]
= 36a2x2 – 25b2y2
(iii) \(\left(\frac{3}{4} p^2+\frac{2}{3} q^2\right)\left(\frac{3}{4} p^2+\frac{2}{3} q^2\right)\) = \(\left(\frac{3}{4} p^2+\frac{2}{3} q^2\right)^2\)
= \(\left(\frac{3}{4} p^2\right)^2+2\left(\frac{3}{4} p^2\right)\left(\frac{2}{3} q^2\right)+\left(\frac{2}{3} q^2\right)^2\)
[Using (a + b)2 = a2 + 2ab + b2 ]
= \(\frac{9}{16}\) p4 + p2q2 + \(\frac{4}{9}\) q4
(iv) (x + \(\frac{3}{4}\)y – 4z)2 = (x)2 + (\(\frac{3}{4}\)y)2 + (- 4z)2 + 2(x) (\(\frac{3}{4}\)y) + 2 (\(\frac{3}{4}\)y) (- 4z) + 2 (x) (- 4z)
[Using (a + b + c)2 = a2 + b2 + c2 – 2ab + 2bc + 2ca]
= x2 + \(\frac{9}{16}\) y2 + 16z2 + \(\frac{3}{2}\) xy – 6yz – 8xz
(v) (x + 2y) (x – 2y) (x2 + 4y2)
= [(x)2 – (2y)2] (x2 + 4y2)
[Using (a + b) (a – b) = a2 – b2]
= (x2 – 4y2) (x2 + 4y2)
= (x2)2 – (4y2)2
= x4 – 16y4
(vi) \(\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)\left(x^4+\frac{1}{x^4}\right)\)
Question 3.
Express the following as square of a binomial and evaluate for the given values:
(i) 4x2 + 12xy + 9y2 ; x = – 2, y = 2
(ii) \(\frac{4}{25}\) m2 – mn + \(\frac{25}{16}\) n2; m = 5, n = 8
Solution:
(i) 4x2 + 12xy + 9y2 = (2x)2 + 2 × (2x) × (3y) + (3y)2
= (2x + 3y)2
[Using (a + b)2 = a2 + 2ab + b2]
[Putting x = – 2 and y = 2]
= [2 (- 2) + 3 (2)]2
= (- 4 + 6)2
= (2)2
= 4
(ii) \(\frac{4}{25}\) m2 – mn + \(\frac{25}{16}\) n2 = \(\left(\frac{2}{5} m\right)^2-2 \times\left(\frac{2}{5} m\right)\left(\frac{5}{4} n\right)+\left(\frac{5}{4} n\right)^2\)
= \(\left(\frac{2}{5} m-\frac{5}{4} n\right)^2\)
[Using (a – b)2 = a2 – 2ab + b2]
= \(\left(\frac{2}{5} \times 5-\frac{5}{4} \times 8\right)^2\)
[Putting m = 5 and n = 8]
= (2 – 10)2
= (- 8)2 = 64
Question 4.
Simplify the following:
(i) (a + b)2 + (a – b)2
(ii) (3x + 7y)2 – (3x – 7y)2
(iii) (6a + 5b + 4c)2 – (6a – 5b – 4c)2
(iv) (a2 – b2) (a2 + b2) – (a2 – b2)2
Solution:
(i) (a + b)2 + (a – b)2
= a2 + 2ab + b2 + a2 – 2ab + b2
= 2a2 + 2b2
= 2(a2 + b2)
[Using (a + b)2 = a2 + 2ab + b2
and (a – b)2 = a2 – 2ab + b2]
(ii) (3x + 7y)2 – (3x – 7y)2
= [(3x)2 + 2(3x) × (7y) + (7y)2] – [(3x)2 – 2(3x) × (7y) + (7y)2]
[Using (a + b)2 = a2 + 2ab + b2
and (a – b)2 = a2 – 2ab + b2]
= (9x2 + 42xy + 49y2) – (9x2 – 42x + 49y2)
= 9x2 + 42xy + 49y2 – 9x2 + 42xy – 49y2
= 42xy + 42xy
= 84xy
(iii) (6a + 56 + 4c)2 – (6a – 56 – 4c)2
= [(6a)2 + (5b)2 + (4c)2 + 2 (6a) (5b) + 2 (5b) (4c) + 2 (6a) (4c)]
– [(6a)2 + (- 5b)2 + (- 4c)2 + 2(6a)(- 5b) + 2 (- 5b) (- 4c) + 2 (6a) (- 4c)]
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= (36a2 + 25b2 + 16c2 + 60b + 40bc + 48ac) – (36a2 + 25b2 + 16c2 – 60ab + 40bc – 48ac)
= 36a2 + 25b2 + 16c2 + 60ab + 40bc + 48ac – 36a2 – 25b2 – 16c2 + 60ab – 40bc + 48ac
= 120ab + 96ac
= 24a (5b + 4c)
(iv) (a2 – b2) (a2 + b2) – (a2 – b2)2
= [(a2)2 – (b2)2] – [(a2)2 – 2a2b2 + (b2)2]
= (a4 – b4) – (a4 + 2a2b2 – b4)
[Using (a + b) (a – b) = a2 – b2 and (a – b)2 = a2 – 2ab + b2]
= a4 – b4 – a4 + 2a2b2 – b4
= 2a2b2 – 2b4
= 2b2 (a2 – b2)
= 2b2 (a + b) (a – b)
Question 5.
Evaluate the following by using a suitable identity.
(i) (399)2
(ii) (62)2
(iii) 103 × 97
(iv) (10.1)2
(v) (85)2 – (75)2
(vi) 291 × 309
Solution:
(i) (399)2 = (400 – 1)2
= (400)2 – 2 × 400 × 1 + (1)2
[Using (a – b)2 = a2 – 2 ab + b2]
= 160000 – 800 + 1
= 159201
(ii) (62)2 = (60 + 2)2
= (60)2 + 2 × 60 × 2 + (2)2
[Using (a + b)2 = a2 + 2 ab + b2]
= 3600 + 240 + 4
= 3844
(iii) 103 × 97 = (100 + 3) (100 – 3)
= (100)2 – (3)2
[Using (a + b)(a – b) = (a2 – b2)]
= 10000 – 9
= 9991
(iv) (10.1)2 = (10 + 0.1)2
= (10)2 + 2 × 10 × 0.1 + (0.1)2
[Using (a + b)2 = a2 + 2ab + b2]
= 100 + 2 + 0.01
= 101.99
(v) (85)2 – (75)2 = (85 + 75) (85 – 75)
[Using a2 – b2 = (a + b) (a – b)]
= 160 × 10
= 1600
(vi) 291 × 309 = (300 – 9) (300 + 9)
= (300)2 – (9)2
[Using (a – b)(a + b) = a2 – b2]
= 90000 – 81
= 89919
Question 6.
Evaluate x if
(i) 36x = (78)2 – (42)2
(ii) 6.2x = 8.1 x 8.1 – 1.9 x 1.9
Solution:
(i) 36x = (78)2 – (42)2
⇒ 36x = (78 + 42) (78 – 42)
[Using a2 – b2 = (a + b) (a – b)]
⇒ 36x = 120 × 36
⇒ x = \(\frac{10 \times 6.2}{6.2}\) = 120
(ii) 6.2x = 8.1 × 8.1 – 1.9 × 1.9
⇒ 6.2x = (8.1)2 – (1.9)2
⇒ 6.2x = (8.1 + 1.9) (8.1 + 1.9)
[Using a2 – b2 = (a + b) (a – b)]
⇒ 6.2x = 10 × 6.2
⇒ x = \(\frac{10 \times 6.2}{6.2}\) = 10
Question 7.
Factorize the following expressions:
(i) z2 – 4z – 77
(ii) x2 + 25x + 144
(iii) x2 + 5x – 104
(iv) 49x2 – 64
(v) x2 – 1 – 2y – y2
(vi) 4(x – y)2 – 12(x – y) + 9
(vii) x2 + y – xy – x
(viii) 81(x + 1)2 + 90 (x + 1) (y + 2) + 25 (y + 2)2
(ix) x2 + y2 + \(\frac{z^2}{4}\) + 2xy – yz – zx
Solution:
(i) z2 – 4z – 77 = z2 – 11z + 7z – 77
[Splitting the middle term]
= z (z – 11) + 1 (z – 11)
= (z – 11) (z + 7)
(ii) x2 + 25x + 144 = x2 + 16x + 9x + 144
[Splitting the middle term]
= x (x + 16) + 9 (x + 16)
= (x + 9) (x + 16)
(iii) x2 + 5x – 104 = x2 + 13x – 8x – 104
= x (x + 13) – 8 (x + 13)
= (x + 13) (x – 9)
(iv) 49x2 – 64 = (7x)2 – (8)2
[Using a2 – b2 = (a + b) (a – b)]
= (7x + 8) (7x – 8)
(v) x2 – 1 – 2y – y2
= x2 – (1 + 2y + y2)
= x2 – (1 + y)2
[Using a2 – b2 = (a + b) (a – b)]
= (x + 1 + y) (x – 1 – y)
(vi) 4 (x – y)2 – 12 (x – y) + 9
[Putting x – y = t]
= 4t2 – 12t + 9
= (2t)2 – 2 (2t) (3) + (3)2
[Using a2 – 2ab + b2 = (a – b)2]
= (2t – 3)2
[Putting t = (x – y)]
= [2 (x – y) – 3]2
= (2x – 2y – 3)2
(vii) x2 + y – xy – x = (x2 – xy) – (x – y)
= x(x – y) – 1 (x – y)
= (x – y) (x – 1)
(viii) 81 (x + 1)2 + 90 (x + 1) (y + 2) + 25 (y + 2)2
[Putting x + 1 = p and y + 2 = q]
= 81p2 + 90pq + 25q2
= (9p)2 + 2(9p)(5q) + (5q)2
[Using a2 + 2 ab + b2 = (a + b)2]
= (9p + 5q)2
(ix) x2 + y2 + – + 2xy – yz – zx
= (x)2 + (y)2 + (\(-\frac{z}{2}\))2 + 2(x)(y) + 2 (y) (\(-\frac{z}{2}\)) + 2 (x) (\(-\frac{z}{2}\))
= (x + y + \(-\frac{z}{2}\))2
[Using a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2]
Question 8.
If x + \(\frac{1}{x}\) = 8, find the value of x2 + \(\frac{1}{x^2}\).
Solution:
x + \(\frac{1}{x}\) = 8,
Squaring both sides, we get
(x + \(\frac{1}{x}\))2 = (8)2
x2 + \(\frac{1}{x^2}\) + 2 (x) (\(\frac{1}{x}\)) = 64
x2 + \(\frac{1}{x^2}\) + 2 = 64
x2 + \(\frac{1}{x^2}\) = 64 – 2 = 62.
Hence x2 + \(\frac{1}{x^2}\) = 62.
Question 9.
If a – \(\frac{1}{a}\) = 5, find the value of a2 + \(\frac{1}{a^2}\).
Solution:
a – \(\frac{1}{a}\) = 5
Squaring both sides, we get
⇒ (a – \(\frac{1}{a}\))2 = (5)2
⇒ a2 + \(\frac{1}{a^2}\) – 2 = 25
⇒ a2 + \(\frac{1}{a^2}\) = 25 – 2 = 23.
Question 10.
If a2 + \(\frac{1}{a^2}\) = 23, find the value of (a + \(\frac{1}{a}\)).
Solution:
(a + \(\frac{1}{a}\))2 = a2 + \(\frac{1}{a^2}\) + 2a . \(\frac{1}{a}\)
= 23 + 2
= 25
∴ a + \(\frac{1}{a}\) = √25 = 5.
Question 11.
If x2 + \(\frac{1}{x^2}\) = 51, find the value of (x – \(\frac{1}{x}\)).
Solution:
(x – \(\frac{1}{x}\))2 = x2 + \(\frac{1}{x^2}\) – 2 . x . \(\frac{1}{x}\)
= x2 + \(\frac{1}{x^2}\) – 2
= 51 – 2 = 49
∴ x – \(\frac{1}{x}\) = √49 = 7
Question 12.
Simplify: \(\frac{2.3 \times 2.3-0.3 \times 0.3}{2.3 \times 2.3-2 \times 2.3 \times 0.3+0.3 \times 0.3}\)
Solution:
\(\frac{2.3 \times 2.3-0.3 \times 0.3}{2.3 \times 2.3-2 \times 2.3 \times 0.3+0.3 \times 0.3}\) = \(\frac{x \times x-y \times y}{x \times x-2 x \times y+y \times y}\)
[Putting 2.3 = x and 0.3 = y]
= \(\frac{x^2-y^2}{x^2-2 x y+y^2}\)
= \(\frac{(x+y)(x-y)}{(x-y)^2}\)
= \(\frac{x+y}{x-y}\)
[Putting x = 2.3 and y = 0.3]
= \(\frac{2.3+0.3}{2.3-0.3}\)
= \(\frac{2.6}{2}\)
= 1.3
Question 13.
If 3x – 4y = 10 and xy = – 1, find the value of 9x2 + 16y2.
Solution:
Given: 3x – 4y = 10 and xy = – 1
∵ (3x – 4y)2 = (3x)2 – 2.3x . 4y + (4y)2
= 9x2 – 24xy + 16y2
(3x – 4y)2 = 9x2 + 16y2 – 24xy
(10)2 = 9x2 + 16y2 – 24 × (- 1)
100 = 9x2 + 16y2 + 24
∴ 9x2 + 16y2 = 100 – 24 = 76.
Question 14.
If 5x – 2y = 7 and xy = 2, find the value of (5x + 2y)2.
Solution:
(5x + 2y)2 = (5x – 2y)2 + 4(5x)(2y)
[Using (x + y)2 = (x – y)2 + 4xy]
= (5x – 2y)2 + 40xy
= (7)2 + 40(2)
= 49+ 80 = 129
Question 15.
If a + \(\frac{1}{a}\) = \(\frac{17}{4}\), find the value of (a – \(\frac{1}{a}\)).
Solution:
Given: a + \(\frac{1}{a}\) = \(\frac{17}{4}\)
∵ \(\left(a-\frac{1}{a}\right)^2=\left(a+\frac{1}{a}\right)^2-4 a \cdot \frac{1}{a}\)
= \(\left(\frac{17}{4}\right)^2\) – 4
= \(\frac{289}{16}\) – 4
= \(\frac{289-64}{16}\)
= \(\frac{225}{16}\)
∴ a – \(\frac{1}{a}\) = \(\sqrt{\frac{225}{16}}\)
= \(\frac{15}{4}\)
DAV Class 8 Maths Chapter 7 HOTS
Question 1.
The sum of (x + 3) observations is (x4 – 81). Find the mean of the observations.
Solution:
Mean Observations = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
= \(\frac{x^4-81}{x+3}\)
= \(\frac{\left(x^2\right)^2-(9)^2}{x+3}\)
= \(\frac{\left(x^2-9\right)\left(x^2+9\right)}{x+3}\)
= \(\frac{\left\{(x)^2-(3)^2\right\}\left(x^2+9\right)}{(x+3)}\)
= \(\frac{(x+3)(x-3)\left(x^2+9\right)}{(x+3)}\)
= (x – 3) (x2 + 9)
Question 2.
The area of a circle is given by the expression (πr2 + 10πx + 25π). Find the radius of the circle.
Solution:
Given: Area of a circle = πx2 + 10πx + 25π
= π (x2 + 10x + 25)
= π (x + 5)2
But, area of a circle = πr2
where r is the radius of the circle.
Thus, πr2 = π (x + 5)2
⇒ r2 = (x + 5)2
∴ r = (x + 5) units.
Additional Questions:
Question 1.
Simplify the following using suitable identities.
(i) (2a2 + 3b)2
(ii) (3x – \(\frac{5}{y}\))2
(iii) (3x2y + z)2
Solution:
(i) (2a2 + 3b)2 = (2a2)2 + 2 (2a2) (3b) + (3b)2
= 4a4 + 12a2b + 9b2
[Using (a + b)2 = a2 + 2ab + b2]
(ii) (3x – \(\frac{5}{y}\))2 = (3x)2 – 2 (3x) (\(\frac{5}{y}\)) + (\(\frac{5}{y}\))2
[Using (a – b)2 = a2 – 2ab + b2]
= 9x2 – 30\(\frac{x}{y}\) + \(\frac{25}{y^2}\)
(iii) (3x2y + z)2 = (3x2y)2 – 2(3x2y) (z) + (z)2
= 9x4y2 + 6x2yz + z2
Question 2.
Simplify the following using suitable identities:
(i) \(\left(\frac{3}{5} x+\frac{5}{6} y\right)^2\)
(ii) \(\left(\frac{2 x}{3}-\frac{3}{y}\right)^2\)
(iii) \(\left(\frac{2 x}{3}-\frac{3}{y}\right)^2\)
Solution:
(i) \(\left(\frac{3}{5} x+\frac{5}{6} y\right)^2=\left(\frac{3}{5} x\right)^2+2\left(\frac{3}{5} x\right)\left(\frac{5}{6} y\right)+\left(\frac{5}{6} y\right)^2\)
= \(\frac{9}{25} x^2+x y+\frac{25}{36} y^2\)
[Using (a + b)2 = a2 + 2ab + b2]
(ii) \(\left(\frac{2 x}{3}-\frac{3}{y}\right)^2=\left(\frac{2 x}{3}\right)^2-2\left(\frac{2 x}{3}\right)\left(\frac{3}{y}\right)+\left(\frac{3}{y}\right)^2\)
= \(\frac{4 x^2}{9}-4 x y+\frac{9}{y^2}\)
[Using (a – b)2 = a2 – 2ab + b2]
(iii) \(\left(6-\frac{5}{y}\right)^2=(6)^2-2(6)\left(\frac{5}{y}\right)+\left(\frac{5}{y}\right)^2\)
= \(36-\frac{60}{y}+\frac{25}{y^2}\)
[Using (a – b)2 = a2 – 2ab + b2]
Question 3.
Simplify the following using suitable identities:
(i) \(\left(\frac{2 x}{3}+\frac{3}{y}-2\right)^2\)
(ii) \(\left(2 x+\frac{3}{x}-1\right)^2\)
Solution:
\(\left(\frac{2 x}{3}+\frac{3}{y}-2\right)^2\) = \(\left(\frac{2 x}{3}\right)^2+\left(\frac{3}{y}\right)^2+(-2)^2+2\left(\frac{2 x}{3}\right)\left(\frac{3}{y}\right)+2\left(\frac{3}{y}\right)(-2)+2\left(\frac{2 x}{3}\right)(-2)\)
= \(\frac{4}{9} x^2+\frac{9}{y^2}+4+4 x y-\frac{12}{y}-\frac{8}{3} x\)
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
(ii) \(\left(2 x+\frac{3}{x}-1\right)^2\) = (2x)2 + (\(\frac{3}{x}\))2 + (- 1)2 + 2 (2x) (\(\frac{3}{x}\)) + 2 (\(\frac{3}{x}\)) (- 1) + 2 (2x) (- 1)
= 4x2 + \(\frac{9}{x^2}\) + 1 + 12 – \(\frac{6}{x}\) – 4x
[Using (a + b + c)2 = a2+ b2 + c2 + 2ab + 2bc + 2ca]
= 4x2 + \(\frac{9}{x^2}\) + 13 – \(\frac{6}{x}\) – 4x
Question 4.
Simplify the following using the suitable identities:
(i) \(\left(2 x^2+\frac{3}{x^2}\right)\left(2 x^2-\frac{3}{x^2}\right)\)
(ii) \(\left(\frac{9 x^2}{2}-\frac{3}{4 x^2}\right)\left(\frac{9 x^2}{2}+\frac{3}{4 x^2}\right)\)
Solution:
(i) \(\left(2 x^2+\frac{3}{x^2}\right)\left(2 x^2-\frac{3}{x^2}\right)\) = (2x2)2 – \(\left(\frac{3}{x^2}\right)^2\)
[Using a2 – b2 = (a + b) (a – b)]
= 4x4 – \(\frac{9}{x^4}\)
(ii) \(\left(\frac{9 x^2}{2}-\frac{3}{4 x^2}\right)\left(\frac{9 x^2}{2}+\frac{3}{4 x^2}\right)\) = \(\left(\frac{9 x^2}{2}\right)^2-\left(\frac{3}{4 x^2}\right)^2\)
[Using a2 – b2 = (a + b) (a – b)]
= \(\frac{81 x^4}{4}-\frac{9}{16 x^4}\)
Question 5.
If (3a + 4b) = 16 and ab = 4, find the value of (9a2 + 16b2).
Solution:
3a + 4b = 16
Squaring both sides, (3a + 46 )2 = (16)2
⇒ (3a)2 + 2 × 3a × 4b + (4b)2 = 256
⇒ 9a2 + 24ab + 16b2 = 256
⇒ 9a2 + 16b2 + 24 × 4 = 256
⇒ 9a2 + 16b2 + 96 = 256
⇒ 9a2 + 16b2 = 256 – 96
⇒ 9a2 + 16b2 = 160
Question 6.
If (x + \(\frac{1}{x}\)) = 5 find the values of
(i) (x2 + \(\frac{1}{x^2}\))
(ii) (x4 + \(\frac{1}{x^4}\))
Solution:
(i) x + \(\frac{1}{x}\) = 5 x
Squaring both sides,
(x + \(\frac{1}{x}\))2 = (5)2
x2 + \(\frac{1}{x^2}\) + 2 × x × \(\frac{1}{x}\) = 25
x2 + \(\frac{1}{x^2}\) + 2 = 25
x2 + \(\frac{1}{x^2}\) = 25 – 2 = 23
(ii) x2 + \(\frac{1}{x^2}\)
Squaring both sides,
(x2 + \(\frac{1}{x^2}\))2 = (23)2
⇒ x4 + \(\frac{1}{x^4}\) + 2 × x2 × \(\frac{1}{x^2}\) = 529
⇒ x4 + \(\frac{1}{x^4}\) + 2 = 529
⇒ x4 + \(\frac{1}{x^4}\) = 529 – 2 = 527
Question 7.
If (x + \(\frac{1}{x}\)) = 6, find the value of x2 – \(\frac{1}{x^2}\).
Solution:
(x + \(\frac{1}{x}\))2 = (x – \(\frac{1}{x}\))2 + 4
⇒ (6)2 = (x – \(\frac{1}{x}\))2 + 4
⇒ 36 = (x – \(\frac{1}{x}\))2 + 4
⇒ (x – \(\frac{1}{x}\))2 = 36 – 4 = 32
x – \(\frac{1}{x}\) = √32
= ± 4√2
Now x2 – \(\frac{1}{x^2}\) = (x + \(\frac{1}{x}\)) (x – \(\frac{1}{x}\))
= 6 × (± 4√2)
= ± 24√2
Question 8.
If (4x + 3y) = 10 and xy = 4, find the value of (16x2 + 9y2).
Solution:
(4x + 3y)2 = (4x)2 + 2 × 4x × 3y + (3y)2
⇒ (4x + 3y)2 = 16x2 + 24xy + 9y2
⇒ (10)2 = 16x2 + 9y2 + 24 x 4
⇒ 100 = 16x2 + 9y2 + 96
∴ 16x2 + 9y2 = 100 – 96 = 4.
Question 9.
If (x2 + \(\frac{1}{25 x^2}\)) = 8\(\frac{3}{5}\), find the value of (x2 – \(\frac{1}{25 x^2}\)).
Solution:
(x2 – \(\frac{1}{25 x^2}\))2 = (x2 + \(\frac{1}{25 x^2}\))2 – 4 × x2 × \(\frac{1}{25 x^2}\)
= \(\left(\frac{43}{5}\right)^2-\frac{4}{25}\)
= \(\frac{1849}{25}-\frac{4}{25}\)
= \(\frac{1845}{25}\)
∴ (x2 + \(\frac{1}{25 x^2}\)) = ± \(\frac{3}{5} \sqrt{205}\)
Question 10.
Factorize: x2 + xy – 2xz – 2yz.
Solution:
x2 + xy – 2xz – 2yz = (x2 – 2xz) + (xy – 2yz)
= x (x – 2z) + y (x – 2z)
= (x – 2z) (x + y)
Question 11.
Show that:
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) \(\left(\frac{4}{3} m-\frac{3}{4} n\right)^2\) + 2mn = \(\frac{16}{9}\) m2 + \(\frac{9}{16}\) n2
Solution:
(i) (3x + 7)2 – 84x = [(3x)2 + 2 (3x) (7) + (7)2] – 84x
= 9x2 + 42x + 49 – 84x
= 9x2 – 42x + 49
= (3x)2 – 2 × 3x × 7 + (7)2
= (3x – 7)2
Hence proved.
(ii) \(\left(\frac{4}{3} m-\frac{3}{4} n\right)^2\) + 2mn = \(\left[\left(\frac{4}{3} m\right)^2-2 \times \frac{4}{3} m \times \frac{3}{4} n+\left(\frac{3 x}{4}\right)^2\right]\) + 2mn
= \(\frac{16}{9}\) m2 – 2mn + \(\frac{9}{16}\) n2
= \(\frac{16}{9}\) m2 + \(\frac{9}{16}\) n2
Hence Proved.
Question 12.
Evaluate:
(i) 9832 – 172
(ii) 194 × 206
Solution:
(i) 9832 – 172 = (983 + 17) (983 – 17)
[Using a2 – b2 = (a + b) (a – b)]
= 1000 × 866
= 866000
(ii) 194 × 206 = (200 – 6) (200 + 6)
= (200)2 – (6)2
[Using (a + b) (a – b) = a2 – b2]
= 40000 – 36
= 39964
Question 13.
Which of the following identities is correct?
(a) (x + a)(x + b) = x2 – (a + b)x + ab
(b) (x – a)(x + b) = x2 + (a + b)x + ab
(c) (x + a)(x – b) = x2 – (a + b)x – ab
(d) (x + a)(x + b) = x2 + (a + b)x + ab
Solution:
(x + a)(x + b) = x2 + (a + b)x + ab
So identity (d) is correct.
Question 14.
Factorize: p2 – pq – 6q2.
Solution:
p2 – pq – 6q2 = p2 – 3pq + 2pq – 6q2
= p (p – 3q) + 2q (p – 3q)
= (p – 3q) (p + 2q)
Question 15.
Factorize: 2x2 + 5y2 + z2 – 2√10 xy – 2√5 yz + 2√2 zx.
Solution:
2x2 + 5y2 + z2 – 2√10 xy – 2√5 yz + 2√2 zx
= (√2x)2 + (- √5 y)2 + (z)2 + 2 × √2x × (- √5y) + 2(- √5y)(z) + 2 (√2 x) (z)
= (√2 x – √5 y + z)2
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]
= (√2 x – √5 y + z) (√2 x – √5 y + z)
Question 16.
If (5x – 2y) = 7 and xy = 2, then find the value of (5x + 2y)2.
Solution:
(5x + 2y)2 = (5x – 2y)2 + 4 (5x) (2y)
= (7)2 + 40 × 2
= 49 + 80
= 129
Question 17.
Factorize: x2 – 1 – 2y – y2.
Solution:
x2 – 1 – 2y – y2 = x2 – (1 + 2y + y2)
= (x)2 – (1 + y)2
= (x + 1 + y) (x – 1 – y)
Multiple Choice Questions:
Question 1.
Using identity, the value of 501 × 502 is
(a) 251502
(b) 255102
(c) 265102
(d) 251052
Solution:
(a) 251502
Question 2.
Factors of x2 – 11x + 30 are:
(a) (x + 6) (x – 5)
(b) (x – 5) (x – 6)
(c) (x + 5) (x + 6)
(d) None of these
Solution:
(b) (x – 5) (x – 6)
Question 3.
If a – \(\frac{1}{a}\) = 5, then the value of a2 + \(\frac{1}{a^2}\) is
(a) 23
(b) 25
(c) 27
(d) 24
Solution:
(c) 27
Question 4.
If (5x – 2y) = 7 and xy = 3, then the value of (5x + 2y)2
(a) 169
(b) 196
(c) 204
(d) 206
Solution:
(a) 169
Question 5.
The value of \(\frac{x^4-1}{x^2-1}\) is
(a) x4 + 1
(b) x2 + 1
(c) x2 – 1
(d) None of these
Solution:
(b) x2 + 1
Question 6.
The value of \(\frac{x^2-x-42}{x-7}\) is
(a) x – 7
(b) x + 7
(c) x – 6
(d) x + 6
Solution:
(d) x + 6
Question 7.
If (x2 + \(\frac{1}{x^2}\)) = 27, then the value of (x – \(\frac{1}{x}\)) is
(a) ± 5
(b) – 5
(c) 5
(d) √5
Solution:
(a) ± 5
Question 8.
The value of (1008)2 is
(a) 1016064
(b) 1014064
(c) 1104064
(d) 1014046
Solution:
(a) 1016064
Question 9.
If a = \(\frac{1}{a-5}\) where a ≠ 5, the value of (a – \(\frac{1}{a}\)) is
(a) 4
(b) 3
(c) 5
(d) √5
Solution:
(c) 5
Question 10.
The value of (9.98)2 is
(a) 9.9602
(b) 9.9604
(c) 9.6904
(d) 9.6604.
Solution:
(b) 9.9604