Class 7 Math Solution WBBSE Algebra Chapter 5 Algebraic Formula Exercise 5 Solved Problems
Necessary Formula (Identities)
1. (a + b)2 = a2 + 2ab+ b2
⇒ (a + b)2 = (a – b) 2 + 4ab
2. (a – b) 2 = a2 – 2ab+b2
⇒ (a – b) 2 = (a + b) 2 – 4ab
3. (a+b+c) 2 = a2 + b2 + c2 + 2ab+ 2bc + 2ca
4. a2+ b2 = (a + b) 2 – 2ab
⇒= (a – b) 2+2ab
5. a2– b2 = (a + b)(a − b)
6. 2(a2 + b2) = (a + b) 2 + (a – b) 2
7. 4ab = (a + b) 2– (a – b) 2
8. ab=(a+b) 2 – (a=b) 2
Proof:
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1. (a + b) 2= (a + b)(a + b).
= a(a + b) + b(a + b)
= a2+ ab + ab + b2
= a2+2ab+ b2
= (a2 – 2ab+b2) + 4ab
= (a – b) 2 + 4ab
2. (a – b) 2 = (a – b)(a – b)
= a(a – b) – b(a – b)
= a2 – ab – ab + b2
= a2 – 2ab+b2
= (a2+2ab+ b2) – 4ab
= (a + b) 2 – 4ab
3. (a + b + c) 2 = {(a + b) + c)} 2
= (a + b) 2 + 2(a + b)c + c2
= a 2 + 2ab+ b 2 + 2ac + 2bc + c2 a 2+ b 2 + c 2+2ab+ 2bc + 2ca
= a2 + b2 = (a 2+2ab+b 2) – 2ab
= (a + b) 2 – 2ab
4. a2+ b2 = (a 2-2ab+b 2)+2ab
= (a – b)2 + 2ab
5. a2 – b2 = a2 + ab – ab-b2
= a(a + b) – b(a + b)
= (a + b)(a – b)
6. 2(a 2+ b 2)= a2 + b2+ a2+ b2
= (a 2+2ab+b 2) + (a 2 – 2ab+b2)
= (a + b) 2 + (a – b)2
7. 4ab = 2ab+2ab
= (a2+2ab+b2) – a2+2ab-b2
= (a2+2ab+b2) + (a2– 2ab+b2)
= (a + b) 2 – (a – b)2
Class 7 Math Solution WBBSE
8. \(a b=\frac{4 a b}{4}=\frac{(a+b)^2-(a-b)^2}{4}\)
= \(\frac{(a+b)^2}{4}-\frac{(a-b)^2}{4}\)
= \(\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\)
Question 1. Choose the correct answer
1. If (x+8)2 = x2 + 16x + K, then the value of K is
1. 16
2. 64
3. 8
4. None of these
Solution:
Given
(x+8)2 = x2 + 16x + K
⇒ x2 + 2. x. 8+ (8)2 = x2 + 16x + K
⇒ 64 K
So the correct answer is 2. 64
2. For which value or values of K, will the expression p2 + pk + \(\frac{1}{25}\) be a perfect square.
1. \(\frac{1}{5}\)
2. \(\frac{1}{25}\)
3. \(\pm \frac{2}{5}\)
4. \(\pm \frac{1}{5}\)
Solution:
Given
p2 + pk + \(\frac{1}{25}\)
So the correct answer is 3. \(\pm \frac{2}{5}\)
3. If a = \(\frac{1}{8}\) then the value of (64a2+ 16a+ 1) is
1. 8
2. 0
3.16
4. 4
Solution:
Given
a = \(\frac{1}{8}\)
64a2 + 16a+ 1
= (8a)2 + 2.8a.1 + (1)2
= (8a+ 1)2
= \(\left(8 \times \frac{1}{8}+1\right)^2\)
[Putting the value of a]
= (1 + 1)2 = (2)2
= 4
⇒ So the correct answer is 4. 4
Wbbse Class 7 Maths Solutions
4. If 3x + \(\frac{1}{5 x}\) =6, then the value of \(25 x^2+\frac{1}{9 x^2}\)
1. 96
2. \(96 \frac{2}{3}\)
3. \(103 \frac{2}{3}\)
4. None of these
Solution:
Given
3x+ \(\frac{1}{5 x}\) = 6
⇒ So the correct answer is 2. \(96 \frac{2}{3}\)
Wbbse Class 7 Maths Solutions
5. If a + b = 7 and a b = 3, then the value of ab is
1. 21
2. 10
3. 58
4. 40
Solution:
Given
a + b = 7, a b = 3
ab = \(\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\)
= \(\left(\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
= \(\begin{aligned}
& =\frac{49}{4}-\frac{9}{4} \\
& =\frac{40}{4}=10
\end{aligned}\)
⇒ So the correct answer is 2. 10
Question 2. Write ‘true’ or ‘false’
1. If (xy)2= (96y+ y)2, then the value of x is 3.
Solution :
Given
(xy)2 = 96y+ y2
⇒ (x-y)2 = (3)2-2.3.y + y2
⇒ (x-y)2=(3-y)2
⇒ x – y = 3-y
⇒x=3-y+y
⇒ x = 3
⇒ So the statement is true.
2. The value of 1015 x 985 is 999085
Solution:
1015 x 985
= (1000+15) (1000-15)
= (1000)2– (15)2
= 1000000 – 225
= 999775
⇒ So the statement is false.
Wbbse Class 7 Maths Solutions
3. If (a-5)2 + (b + 3)2= 0, then the value (a + b) is 2
Solution:
Given
If (a-5)2 + (b + 3)2= 0
If the sum of two or more two square is zero then the value of each square will be zero.
(a-5)2 = 0
⇒ a 5=0
⇒ a = 5
(b + 3)2 = 0
⇒b+3=0
⇒ b = -3
a+b=5-3=2.
⇒ So the statement is true.
Question 3. Fill in the blanks
1. The value of (0.75 x 0.75+1.5 x 0.25 +0.25 x 0.25 is) _______
Solution: 0.75 x 0.75 + 1.5 x 0.25 + 0.25 × 0.25
= (0-75)2 + 2 x 0.75 x 0.25 + (0.25)2
= (0·75 + 0.25)2
= (1.00)2
= 1
2. If a + b = 5 and ab= 6, then the value of (a – b) is _____
Solution:
Given
If a + b = 5 and ab= 6
(a – b)2 = (a + b)2 – 4ab
= (5)2 – 4 x 6
= 25-24
= 1
⇒ a-b=±√l=±1
3. If a + b + c = 7 and ab+be+ca = -5, then the value of (a2+ b2+ c2) is _____
Solution:
Given
If a + b + c = 7 and ab+be+ca = -5
a2 + b2 + c2 + 2 (ab + bc + ca) = (a + b + c)2
⇒ a2 + b2 + c2 + 2 x (-5) = (7)2
⇒ a2 + b2 + c2 = 49+ 10 = 59
Question 4. Find the square of the algebraic expression given below:
1. 4x+5y
2. 2a+3b-4c
3. a +2b3c4d
4. 999
5. 1005
Solution:
Given
1. (4x+5y)2
= (4x)2 + 2.4x.5y + (5y)2
= 16x2 + 40xy + 25y2
(4x+5y)2 = 16x2 + 40xy + 25y2
2. (2a+3b – 4c)2
= {(2a + 3b) – 4c)2
= (2a+3b)2 – 2 x (2a + 3b) x 4c + (4c)2
= (2a)2 + 2 x 2a x 3b + (3b)2 – 8c(2a + 3b) + 16c2
= 4a2 + 12ab+9b2 -16ac – 24bc + 16c2
= 4a2 + 9b2 + 16c2 + 12ab – 16ac – 24bc.
⇒ (2a + 3b) – 4c)2 = 4a2 + 9b2 + 16c2 + 12ab – 16ac – 24bc.
3. (a + 2b – 3c + 4d)2
= {(a + 2b) – (3c – 4d)}2
= (a + 2b)2– 2(a + 2b)(3c4d) + (3c-4d)2
= a2+2.a.2b + (2b)2 – 2(3ac-4ad + 6bc-8bd) + (3c)2 – 2.3c.4d + (4d)2
= a2+4ab + 4b2 – 6ac + 8ad – 12bc + 16bd + 9c2 – 24cd + 16d2
= a2 + 4b2 + 9c2 + 16d2 + 4ab – 6ac + 8ad -16bd – 24cd.
⇒ {(a + 2b) – (3c – 4d)}2 = a2 + 4b2 + 9c2 + 16d2 + 4ab – 6ac + 8ad -16bd – 24cd.
4. (999)2
= (1000 – 1)2
= (1000)2-2 × 1000 × 1 + (1)2
= 1000000-2000 + 1
= 998001
⇒ (999)2= 998001
5. (1005)2
= (1000 + 5)2
= (1000)2 + 2 x 1000 × 5 + (5)2
= 1000000 + 10000 + 25.
= 1010025
⇒ (1005)2 = 1010025
Wbbse Class 7 Maths Solutions
Question 5. Express x as a difference of two square.
Solution: x = x.1
= \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)
Question 6. Express (8a2+ 50b2) as a sum of two square.
Solution:
Given
8a2+ 50b2
= 2 (4a2 + 25b2)
= 2{(2a)2 + (5b)2}
= (2a + 5b)2 + (2a – 5b)2
Question 7. Find the square (3x+4y) with the help of (a – b)2= a2 – 2ab+ b2
Solution:
(3x+4y)2
= {3x-(-4y)}2
= (3x)2– 2 x 3x(-4y) + (-4y)2
= 9x2 + 24xy + 16y2
Question 8. For what values of p; will the expression (9x2 + px + 16) be a perfect square.
Solution :
Given
9x2 + px + 16
=(3x)2+2.3x.\(\frac{p}{6}\)+ \(\left(\frac{p}{6}\right)^2\) – \(\left(\frac{p}{6}\right)^2\) +16
= \(\left(3 x+\frac{p}{6}\right)^2-\frac{p^2}{36}+16\)
The given expression will be perfect square if
\(-\frac{p^2}{36}\)+16=0
\(-\frac{p^2}{36}\) = -16
⇒ p2 = 16 x 36
⇒ p=±\(\sqrt{16 \times 36}\)
⇒p=14×6
⇒p±24.
Wbbse Class 7 Maths Solutions
Question 9. Express the following in the product form
1. 49x4 – 36y4
2. (m + p + q)2 – (m-p-q)2
Solution:
Given
1. 49x4 – 36y4
= (7x2)2– (6y2)2
= (7x2+6y2)(7x2 – 6y2)
49x4 – 36y4 = (7x2+6y2)(7x2 – 6y2)
Given
2. (m + p + q)2– (m-p-q)2
= (2m)(2p+ 2q)
= 2m x 2(p + q)
= 4m (p + q)
⇒ (m + p + q)2– (m-p-q)2 = 4m (p + q)
Question 10. For what value of t, will the expression \(\left(x^2-t x+\frac{1}{4}\right)\) be a perfect square.
Solution: \(x^2-t x+\frac{1}{4}\)
= \(x^2-2 \cdot x \cdot \frac{t}{2}+\left(\frac{t}{2}\right)^2-\left(\frac{t}{2}\right)^2+\frac{1}{4}\)
= \(\left(x-\frac{t}{2}\right)^2-\frac{t^2}{4}+\frac{1}{4}\)
The given expression is a perfect square.
So, \(-\frac{t^2}{4}+\frac{1}{4}=0\)
⇒ \(-\frac{t^2}{4}=-\frac{1}{4}=0\)
⇒ t2 = 1
= t = ± 1
Wbbse Class 7 Maths Solutions
Question 11. Express the following as a perfect square and hence find the values
1. 49a2 – 42ab+9b2 [when a = 1, b = 2]
2. \(\frac{144}{p^2}-\frac{120}{p}+25\) [when p = -3]
Solution: 1. 49a2 – 42ab+9b2
= (7a)2 – 2 x 7a x 3b + (3b)2
= (7a-3b)2
= (7 x 1-3 x 2)2
= (7-6)2
= (1)2 = 1
2. \(\frac{144}{p^2}-\frac{120}{p}+25\)
= \(\left(\frac{12}{p}\right)^2-2 \times \frac{12}{p} \times 5+(5)^2\)
= \(\left(\frac{12}{p}-5\right)^2\)
= \(\left(\frac{12}{-3}-5\right)^2\) [Putting p = -3]
= (-4-9)2
= (-13)2 = 16
Question 13. If \(m-\frac{1}{m-5}=12\) then find the value of \((m-5)^2+\frac{1}{(m-5)^2}\)
Solution:
Given
\(m-\frac{1}{m-5}=12\)
The value of \((m-5)^2+\frac{1}{(m-5)^2}\) = 51
Wbbse Class 7 Maths Solutions
Question 14. If 3x – \(\frac{1}{x}\)=6, then find the value of \(\left(x^2+\frac{1}{9 x^2}\right)\)
Solution:
Given
3x – \(\frac{1}{x}\)=6
Question 15. If a2 + b2= 52 and a – b = 2, then find the value of ab.
Solution:
Given
a2 + b2 = 52
⇒ (a – b)2+2ab = 52
⇒ (2)2+2ab = 52
⇒ 2ab = 52 – 4 = 48
⇒ ab= \(\frac{48}{2}\) =24
The value of ab = 24
Question 16. If a + b = √3 and a b= √2, then find the value of
1.8ab (a2 + b2)
2. \(\frac{3\left(a^2+b^2\right)}{a b}\)
Solution:
1. 8ab (a2 + b2)
= 4ab x 2(a2 + b2)
= {(a + b)2 – (a – b)2} {(a + b)2 + (a – b)2}
= {(V3)2– (√2)2}{(√3)2 + (√2)2} [ Putting a + b = √3 and a b= √2 ]
= (3-2)(3 + 2)
= 1 x 5
= 5
8ab (a2 + b2) = 5
2. \(\frac{3\left(a^2+b^2\right)}{a b}\)
= \(6 \times \frac{2\left(a^2+b^2\right)}{4 a b}\)
= \(6 \times \frac{(a+b)^2+(a-b)^2}{(a+b)^2-(a-b)^2}\)
= \(6 \times \frac{3+2}{3-2}=6 \times \frac{5}{1}\) = 30
\(\frac{3\left(a^2+b^2\right)}{a b}\) = 30
Question 17. Express as the difference of two squares
1. 72
2. (a + 3b)(2a – 5b)
Solution:
1. 72 = 9 x 8
2. (a + 3b)(2a 5b)=
Wbbse Class 7 Maths Solutions
Question 18. If x2 – 3x-1= 0, then find the value of \(\left(x^4+\frac{1}{x^4}\right)\)
Solution:
Given
x2 – 3x-1= 0
The value of \(\left(x^4+\frac{1}{x^4}\right)\) is 119.
Question 19. If x = \(a+\frac{1}{a}\) and y = \(a-\frac{1}{a}\) then find the value of (x2 + y2 – 2x2y2).
Solution:
Given
If x = \(a+\frac{1}{a}\) and y = \(a-\frac{1}{a}\)
The value of (x2 + y2 – 2x2y2) is 16.
Question 20. If \(\frac{a}{b}+\frac{c}{d}=\frac{b}{a}+\frac{d}{c}\) then prove that \(\frac{a^2}{b^2}-\frac{c^2}{d^2}=\frac{d^2}{c^2}-\frac{b^2}{a^2}\)
Solution:
Given
\(\frac{a}{b}+\frac{c}{d}=\frac{b}{a}+\frac{d}{c}\)
Wbbse Class 7 Maths Solutions
Question 21. Show that (a+b) (a2 + b2)(a4 +b4)(a3 +b8 + b8) = \(\frac{a^{16}-b^{16}}{a-b}\)
Solution:
Wbbse Class 7 Maths Solutions
Question 22. If a2 + b2 + c2 = 2(a – b − 1), then find the value of (a + b + c).
Solution:
Given
a2+ b2 + c2 = 2(a – b-1)
⇒ a2 + b2 + c2 = 2a-2b-2
⇒ a2 – 2a + b2 + 2b + c2 + 2 = 0
⇒ (a2 – 2a + 1) + (b2 + 2b + 1) + c2 = 0
⇒(a – 1)2 + (b +1)2 + c2 = 0
If sum of square of two or more than two square is zero then each square will be zero.
(a -1)2 = 0
⇒ a-1=0
⇒ a=1
(b + 1)2 = 0
⇒b+1=0
⇒ b = -1
c2 = 0
⇒ c=0
a+b+c=1=1-1+0=0
The value of (a + b + c) is 0.
Class 7 Math Solution WBBSE
Question 23. Prove that (a + b)4 – (a – b)4= 8ab (a2 + b2).
Solution:
Proof:
(a + b)4 – (a – b)4
= \(\left\{(a+b)^2\right\}^2-\left\{(a-b)^2\right\}^2\)
= \(\left\{(a+b)^2+(a-b)^2\right\}\)\(\left\{(a+b)^2-(a-b)^2\right\}\)
=2(a2+b2) x 4ab
= 8ab(a2+ b2) (Proved)
(a + b)4 – (a – b)4= 8ab (a2 + b2)
Question 24. Multiply (a+b+c) (a – b + c)(b + c – a)(a + b – c).
Solution:
Given
(a+b+c)(ab+c)(b + c a)(a + b – c)
= {(a + c) + b}{(a + c) – b}{b + (ca)} {b (c – a)}
= {(a + c)2 – b2}{b2– (c – a)2}
= (a2+2ac + c2– b2) (b2 – c2 + 2ac – a2)
= {2ac + (a2 – b2 + c2)}{2ac – (a2 – b2 + c2)}
= (2ac)2– (a2 – b2 + c2)2
= 4a2c2 – {(a2 – b2)2 + 2(a2 – b2)c2 + (c2)2}
= 4a2c2 – {a4 – 2a2b2 + b4 + 2a2c2 – 2a2c2 + c4) = 4a2c2-a4 + 2a2b2 – b4 – 2a2c2 + 2b2c2 – c4
= 2a2b2 + 2b2c2 + 2a2c2 – a4– b4 – c4
(a+b+c)(ab+c)(b + c a)(a + b – c) = 2a2b2 + 2b2c2 + 2a2c2 – a4– b4 – c4