Algebra Chapter 5 Algebraic Formula Exercise
Question 1. Choose the correct answer.
1. If (25x² + pxy+ 36y²) is a perfect square then the Value of P is
Solution:
25x² + Pxy + 36y²
⇒ (5x)² + pxy + (6y)²
Now this expression is in the form of (a+b)²= a²+2ab+b²
Here, a = 5x, b = 6y
∴ 2ab = рху
2(5x)(6x) = pxy
2(30xy) = pxy
60xy = pxy
P = 60
∴ The value of the ‘p’ is 60′
Read and Learn More Class 7 Maths Solutions
2. If x² – 12x + 36 = (x+m)² then the value of ‘m’?
Solution:
⇒ x² – 12x+36 = (x+m)²
⇒ x² – 2.6x + (6)² = (x+m)²
⇒ (x-6)² = (x+m)²
Cancel the equal terms on both sides
⇒ (x-6)² = (x+m)²
⇒ x – 6 – x – m = 0
⇒ -6 – m =0
⇒ -(6+m)=0
m + 6 =0
m = -6
∴ x² – 12x+36= (X+m)² The value of m is -6′
Algebra Formulas For Class 7 WBBSE
3. If a+b+c= 6 and a² + b² + c² = 8 then the value of (ab+ bc+ca) is
- 28
- 24
- 14
- 12
Solution:
a+b+c= 6 and a²+ b²+c²=8.
Squaring the equation a+b+c=6;
⇒ (a+b+c)² = 62
⇒ a² + b²+c² + 2(ab+bc+ca) =36
Given that a² + b² + c² = 8 substitute in above expression
⇒ 8 + 2(ab+bc+ca) = 36
⇒ 2(ab+bc+ca)=36-8
⇒ (ab+bc+ca) = \(\frac{28}{2}\)
∴ ab+bc + ca = 14
The value of (ab+bc+ca) is 14
4. If a² – b² = 55 [a>b] then the value of a and b are respectively.
- 8,3
- 11,5
- 3,8
- 5,11
Solution:
a² – b² =55
⇒ (a+b)(a-b) = 55
By checking the options in a given condition.
⇒ 8, 3
a² – b² = 55
(8)² – (3)² = 55
64 – 9 =55
55 = 55
∴ The values of a and b are 8, 3
Class 7 Maths Chapter 5 Solved Exercises
5. If \(x^4+\frac{1}{x^4}=119\) then the value of \(\left(x-\frac{1}{x}\right)\)
- 11
- 121
- 9
- 3
Solution:
⇒ \(x^4+\frac{1}{x^4}=119\)
⇒ \(\left(x^2\right)^2+2 \cdot x^2 \frac{1}{-x^2}+\frac{1}{\left(-x^2\right)^2}=119\)
⇒ \(\left(x^2\right)^2+\frac{1}{\left(-x^2\right)^2}=119+2\)
⇒ \(\left(x^2\right)^2+\frac{1}{\left(x^2\right)^2}=121\)
Applying square root on both sides.
⇒ \(\sqrt{\left(x^2\right)^{x^1}+\left(\frac{1}{x^2}\right)^2}=\sqrt{121}\)
⇒ \(x^2+\frac{1}{x^2}=11\)
⇒ \(x^2+2 x \cdot \frac{1}{-x}+\left(\frac{1}{-x^2}\right)^2=11\)
⇒ \(x^2-\frac{1}{x^2}=11-2\)
⇒ \(x^2-\left(\frac{1}{x}\right)^2=9\)
Applying square root on both sides.
⇒ \(\sqrt{(x)^2-\left(\frac{1}{x}\right)^x}=\sqrt{9}\)
⇒ \(x-\frac{1}{x}=3\)
∴ Option (4) 3 is the correct Answer.
6. If 3a-\(\frac{1}{5a}\) = 12 then the value of (25a² + \(\frac{1}{9a^2}\)) is
- 147\(\frac{1}{3}\)
- 403\(\frac{1}{3}\)
- 7\(\frac{1}{5}\)
- 5\(\frac{1}{7}\)
Solution:
3a – \(\frac{1}{5a}\) Multiplying \(\frac{5}{3}\) on both sides
⇒ \(\frac{5}{3}\left(3 a-\frac{1}{5 a}\right)=14 \times \frac{5}{3}\)
⇒ \(\frac{5}{7} \times 3 a-\frac{8}{3} \times \frac{1}{5 a}=20\)
5a – \(\frac{1}{3 a}=20\)
Squaring on Both sides
⇒ \(\left(5 a-\frac{1}{3 a}\right)^2=(20)^2\)
(5a)\(^2+\left(\frac{1}{3 a}\right)^2-2(5 a) \times\left(\frac{1}{3 a}\right)=400\)
25 \(a^2+\frac{1}{9 a^2}=400+2\left(\frac{5}{3}\right)\)
= \(\frac{3(400)+10}{3}\)
= \(\frac{1200+10}{3}\)
= \(\frac{1210}{3}\)
= \(403 \cdot 33\)
∴ 25 \(a^2+\frac{1}{9 a^2}=403 \cdot \frac{1}{3}\)
Question 2. Write true or false
1. If x² + \(\frac{1}{x^2}\) = 2 then the value of (x-\(\frac{1}{x}\)) is 0
Solution:
x² + \(\frac{1}{x^2}\) -2(x)(\(\frac{1}{x}\))=0
a² + b² -2ab = (a – b)²
(x-\(\frac{1}{x}\))² = 0
(x-\(\frac{1}{x}\)) = 0
∴ True
2. If 3a – 5b then the value of (9a² – 30ab+25b²) is 64
Solution:
9a² – 30ab+ 25b² = 64
(3a)² – 2(30) (5b)+(5b)² = 64
(3a-5b)² = 64
∴ 3a=5b
(5b-5b)² = 64
(0)²=64
∴ False
Class 7 Algebra Problems With Solutions
3. If x² – 6x + 9=0 then the value of (x² + 6x + 9) is 16
Solution:
x² – 6x + 9 = 0
x² – 2·x·3 + (3)² = 0
(x – 3)² = 0
x – 3 =0
∴ x = 3
x² + 6x + 9 = 16
⇒ (3)² + 6(3) + 9 = 16
⇒ 9+18+9 = 16
⇒ 36 = 16
∴ False
Question 3. Fill in the blanks
1. (a+2b)² =(a-2b)² + 8ab
Solution:
(a+b)² = (a−b)² + 4ab
(a+2b)² = (a-2b)² + 4a(2b)
= (a-2b)²+ 8ab
(a+b)² = (a-2b)² + 8ab
2. The value of (7.3 x 7.3 + 2·3 × 2·3 – 7.3 × 4.6) is
Solution:
⇒ 7.3 x 7.3 + 2·3 × 2·3 – 7.3 × 4.6
⇒ 53.29 + 5.29 – 33.58
⇒ 58.58 – 33.58
⇒ 25
∴ (7·3 × 7·3 + 2.3 × 2·3 – 7·3 × 4.6) = 25
3. If (x-y)² = x² – 10x +25 then the value of y is
Solution:
= (x-y)² = x² – 10x + 25
(x-y)² = x² – 2·x·5 + (5)²
∴ The value of ‘y’ is ‘5’
(x-y)² = x² – 2xy + y²
2xy = 2x.5
∴ y =5
Question 4. Find the Square
- 995
- 805
- 2a-3b+4c
Solution:
1. 995 x 995 = 990025
2. 805 x 805 = 648025
3. (2a-3b+4c)²
⇒ (2a+(-3b)+4c)²
⇒ (a+b+c)²= a² + b²+c² + 2ab+ 2bc + 2ca
Here a= 2a, b= -3b, c = 4c
⇒ (2a+(-3b)+4c)² = (2a)² + (-3b)² +(4c)² + 2(2a)(−3b) + 2(3b)(4c) + 2(4c)(2a)
= 4a² + 9b² + 16c² – 12ab – 24bc + 16ac
Class 7 Maths Exam Preparation WBBSE
Question 5. Find the value of 998 × 1002
Solution:
⇒ \(\begin{array}{r}
1002 \\
\times 998 \\
\hline 8016 \\
9018 \times \\
9018 \times \\
\hline 999996 \\
\hline
\end{array}\)
∴ 998 × 1002 = 999996
Question 6. Express (a²+b²) (c²+ d²) as a sum of two square
Solution:
(a²+b²)(c²+d²)
Let us expand it a²c² + a²d² + c²b² + b²d² ⇒ (a²c² + b²d²) + (a²d² + b²c²)
Notice that both pairs have the same structure but with different terms. Each pair is a perfect square of two terms.
⇒ a²d²+ b²c²
⇒ (ad)² + (bc)²
Or,
⇒ (a²c²+b²d²)
⇒ (ac)² + (bd)²
∴ (a² + b²) (c² + d²) can be expressed as the
Sum of two squares: (ac)² +(bd)² + (ad)² + (bc)²
Question 7. If (a-3)² + (b-2)² = 0 then find the value of (a+3)² + (b+2)²
Solution:
(a-3)² + (b-2)² = 0
(a – 3)² =0,
a – 3 = 0
a = 3
and
(b – 2)² =0
b – 2 = 0
b = 2
(a+3)² + (b+2)²
Now Substitute the a and b values in the above equation.
(3+3)²+(2+2)²
(6)² +(4)²
⇒ 36+16
⇒ 52
∴ (a+3)² + (b+2)² = 52
Class 7 Maths Algebra Solutions WBBSE
Question 8. If x² – ax – 1 = 0 then find the value of \(\left(x^4+\frac{1}{x^4}\right)\)
Solution:
x² – ax – 1 = 0
⇒ x² – 1 = ax
⇒ \(\frac{x^2-1}{x}=a, \Rightarrow x-\frac{1}{x}=a\)
squaring on both sides
⇒ \(\left(x-\frac{1}{x}\right)^2=(a)^2\)
⇒ \(x^2-2 \cdot x\left(\frac{1}{x x}\right)+\frac{1}{x^2}=a^2\)
⇒ \(x^2+\frac{1}{x^2}=a^2+2\)
⇒ \(\left(x^2+\frac{1}{x^2}\right)^2=\left(a^2+2\right)^2\) (squaring on both sides)
⇒ \(\left(x^2\right)^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)+\left(\frac{1}{x^2}\right)^2=\left(a^2\right)^2+(2)^2+2\left(a^2\right)(2)\)
⇒ \(\left(x^4+\frac{1}{x^4}\right)=a^4+4+4 a^2-2\)
⇒ \(x^4+\frac{1}{x^4}=a^4+4 a^2+2\)
∴ \(\left(x^4+\frac{1}{x^4}\right)=a^4+4 a^2+2\)
Question 9. If a = 2017, b = 2018, and c = 2019 then Find the value of (a²+b+c²-ab-bc-ca)
Solution:
a = 2017, b = 2018, C = 2019
(a² + b² + c² – ab – bc – ca)
⇒ a² + b² + c² – (ab + bc + ca)
⇒ (2017) + (2018) + (2019)² – {(2017x 2018) + (2018x 2019) + (2019×2019)}
⇒ {(4068289 + 4072324 + 407636)} – {(4070306) + (4074342) + (4072323)}
⇒ 12216974 – 12216971
⇒ 3
∴ (a² + b + c² – ab – bc – ca) = 3
Question 10. If 4x – \(\frac{1}{4x}\) = 16 then find the value of x² + \(\frac{1}{256 x^2}\)
Solution:
4x – \(\frac{1}{4x}\) = 16
Multiply with \(\frac{1}{4}\) on Both sides
⇒ \(\left(4 x-\frac{1}{4 x}\right) \times \frac{1}{4}=46 \times \frac{1}{4}\)
⇒ \(\frac{4 x}{4}-\frac{1}{16 x}=4\)
squaring on Both sides
⇒ \(\left(x-\frac{1}{16 x}\right)^2=(4)^2 \quad(a-b)^2=a^2-2 a b+b^2\)
⇒ \(x^2-2 \cdot x \cdot\left(\frac{1}{16 x}\right)+\left(\frac{1}{16 x}\right)^2=16\)
⇒ \(x^2+\frac{1}{256 x^2}=\frac{16+\frac{1}{8}}{8}\)
⇒ \(x^2+\frac{1}{256 x^2}=\frac{128+1}{8}\)
⇒ \(x^2+\frac{1}{256 x^2}=\frac{129}{8}\)
⇒ \(x^2+\frac{1}{256 x^2}=16.125\)
⇒ \(16 \frac{1}{8}\)
∴ \(x^2+\frac{1}{256 x^2}=16 \cdot \frac{1}{8}\)
Class 7 Maths Algebra Solutions WBBSE
Question 11. If 3(a² + b² + c²) = (a + b + c)² then Find a:b:c
Solution:
3(a² + b² + c²) = (a + b + c)²
3a² + 3b² + 3c² = a² + b² + c² + 2(ab + bc + ca)
3a² + 3b² + 3c² – (a² + b² + c) = 2(ab + bc + ca)
3a² + 3b² + 3c² – a² – b² – c² = 2(ab + bc + ca)
2a² + 2b² + 2c² = 2(ab+bc+ca)
Dividing the equation by ‘2’ on Both sides
a² + b² + c² = ab + bc + ca
a² + b² + c² = ab – bc + ca = 0
Now Let’s Factorize and rearrange.
a² – 2ab + b² + a² – 2ac + c² + b² + 2bc + c² = 0
(a – b)² + (a-c)² + (b – c)² = 0
For the expression to be equal to zero
(a-b)² = 0
a – b = 0
and
(a-c)² = 0
a – c = 0
and
(b-c)² = 0
b – c = 0
This implies. a=b = c
∴ The ratio of a:b:c is 1:1:1
Question 13. Express \(\left(x^2-\frac{1}{y^2}\right)\left(y^2-\frac{1}{x^2}\right)\) as a perfect square
Solution:
⇒ \(\left(x^2-\frac{1}{y^2}\right)\left(y^2-\frac{1}{x^2}\right)\)
⇒ \(x^2\left(y^2-\frac{1}{x^2}\right)-\frac{1}{y^2}\left(y^2-\frac{1}{x^2}\right)\)
⇒ \(x^2 y^2-\frac{x^2}{x^2}-\frac{y^2}{x^2}+\frac{1}{x^2 y^2}\)
⇒ \(x^2 y^2-1-1+\frac{1}{x^2 y^2}\)
⇒ \(x^2 y^2-2+\frac{1}{x^2 y^2}\)
⇒ \(x^2 y^2-2(x y)\left(\frac{1}{x^2 y}\right)+\frac{1}{x^2 y^2}\)
(\(a^2-2 a b+b^2=(a-b)^2\); \(a^2=x^2 x^2\); \(b^2=\frac{1}{x^2 y^2}\))
⇒ \(\left(x y-\frac{1}{x y}\right)^2\)
∴ \(\left(x^2-\frac{1}{y^2}\right)\left(y^2-\frac{1}{x^2}\right)=\left(x y-\frac{1}{x y}\right)^2\)
Question 15. Express the following as a perfect square and hence Find the values
1. \(\frac{169}{a^2}-\frac{104}{a b}+\frac{16}{b^2}\) when a = 13, and b=-4
Solution:
⇒ \(\frac{169}{(13)^2}-\frac{104}{(13)(-4)}+\frac{16}{(-4)^2}\)
⇒ \(\frac{169}{169}-\frac{104}{-52}+\frac{16}{16}\)
⇒ \(1+\frac{104}{52}+1\)
⇒ 1 + 2 + 1
⇒ 4
∴ \(\frac{169}{a^2}-\frac{104}{a b}+\frac{16}{b^2}=4\)
2. \(121 a^2 b^2-66 a b+9\) when a = 1 b = -1
Solution:
⇒ \(121(1)^2(-1)^2-66(1)(-1)+9\)
⇒ 121+66+9
⇒ 196
∴ \(121 a^2 b^2-66 a b+9=196\)
WBBSE Class 7 Algebra Exercise Solutions
Question 16. Multiply: \(\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4+a^2 b^2+b^4\right)\)
Solution:
⇒ \(\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4+a^2 b^2+b^4\right)\)
⇒ \(\left\{\left(a^2+b^2\right)+a b\right\}\left\{\left(a^2+b^2\right)-a b\right\}\left(a^4+a^2 b^2+b^4\right)\)
⇒ \(\left\{\left(a^2+b^2\right)^2-(a b)^2\right\}\left\{a^2+a^2 b^2+b^4\right\}\)
((a+b)(a-b)=\(a^2-b^2\)
(a+b)\(^2=a^2+b^2+2 a b\))
⇒ \(\left.\Rightarrow\left\{\left(a^4+b^4+2 a^2 b^2\right)\right\}-a^2 b^2\right\}\left\{a^4+a^2 b^2+b^4\right\}\)
⇒ \(\left\{\left(a^4+a^2 b^2+b^4\right)\right\}\left\{\left(a^4+a^2 b^2+b^4\right)\right\}\)
⇒ \(\left(a^4+a^2 b^2+b^4\right)^2\)
⇒ \(\left(a^4\right)^2+\left(a^2 b^2\right)^2+\left(b^4\right)^2 \Rightarrow\left(a^4 \times a^4\right)+a^2 b^2 \times a^2 b^2+b^4 \times b^4\)
⇒ \(a^8+a^4 \cdot b^4+b^8\) (\(a^m \cdot a^n=a^{m+n}\))
∴ \(\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4+a^2 b^2+b^4\right)=a^8+a^2 b^4+b^8\)
Question 17. If a + b = 10, and a – b = 4, then find the value of 7ab(a² + b²)
Solution:
a + b = 10
Squaring on Both sides
(a + b)² = (10)²
⇒ a² + b² + 2ab = 100
⇒ a² + b² = (100 – 2ab)
⇒ (a – b)2 2ab = (100 – 2ab)
⇒ (4)² + 2ab = 100 – 2b
⇒ 2ab + 2ab = 100 – 16
⇒ 4ab = 84
ab = \(\frac{84}{4}\)
∴ ab = 21
a² + b² = 100 – 2ab
= 100 – 2(21)
=100 – 42
∴ a² + b² = 58
7ab(a² + b²) = 7(21)(58)
7ab(a²+b²)= 8526
WBBSE Class 7 Algebra Exercise Solutions
Question 18. If \(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=1\) then find the value of \(\left(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\right)\)
Solution:
⇒ \(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=1\)
Adding 3 on Both sides
1 + \(\frac{a}{1-a}+1+\frac{b}{1-b}+1+\frac{c}{1-c}=1+3\)
⇒ \(\frac{1-a+a}{1-a}+\frac{1-b+b}{1-b}+\frac{1-c+c}{1-c}=4\)
⇒ \(\frac{1}{(1-a)}+\frac{1}{(1-b)}+\frac{1}{(1-c)}=4\)
Question 19. (x+y+z) (x-y+z) (x+y-z) (y+z-x)
Solution:
{(x+z)+y}{(x+z)-y}{y-(z-x)}{y+(z-x)} [(a+b)(a−b) = a² – b²]
{(x+z)² + y2²} {(y)² -(z-x)²}
{x²+ z²+2xz-y²} {y²-(z²+x²- 2xz)}
{x²+z²+2xz-y²} {y²-z²-x²+ 2xz}
{(2xz)² + (x² – y+ z²)} {(2xz)-(x²- y²+z²)}
(2xz)² – (x²-y²+z²)²
4x²z² – ((x²- y²)² + 2(x² – y²)z² + (z²)²}
4x²z² – {(x4 +y4 – 2x²y² + 2x²z² – 2y²z² + z4
4x²z² – x4 – y4 + 2x²y² – 2x²z² + 2y²z² – z4
2x²z² + 2 y²z² + 2x²y² =x4 – y4 – z4
∴ (x + y +z) (z – y + z) (x + y – z) (y + z – x) = 2x²z² + 2y²z² + 2x²y² – x4 – y4 – z4