Class-8th Ex.5.1 – Class 7 and 8 Resources Center

Exercise 5.1 – Part 1

Q1. Find the square of the following expressions.

(a) Find the square of \( (3x+4y) \).

\[ (a+b)^2=a^2+2ab+b^2 \] \[ (3x+4y)^2 =(3x)^2+2(3x)(4y)+(4y)^2 \] \[ =9x^2+24xy+16y^2 \]

Answer: \[ \boxed{9x^2+24xy+16y^2} \]

(b) Find the square of \( (9x+8y) \).

\[ (9x+8y)^2 =(9x)^2+2(9x)(8y)+(8y)^2 \] \[ =81x^2+144xy+64y^2 \]

Answer: \[ \boxed{81x^2+144xy+64y^2} \]

(c) Find the square of \( (4x-5y) \).

\[ (a-b)^2=a^2-2ab+b^2 \] \[ (4x-5y)^2 =(4x)^2-2(4x)(5y)+(5y)^2 \] \[ =16x^2-40xy+25y^2 \]

Answer: \[ \boxed{16x^2-40xy+25y^2} \]

(d) Find the square of \( (7x-3y) \).

\[ (7x-3y)^2 =(7x)^2-2(7x)(3y)+(3y)^2 \] \[ =49x^2-42xy+9y^2 \]

Answer: \[ \boxed{49x^2-42xy+9y^2} \]

(e) Find the square of \( (x^2+y^2) \).

\[ (x^2+y^2)^2 =(x^2)^2+2(x^2)(y^2)+(y^2)^2 \] \[ =x^4+2x^2y^2+y^4 \]

Answer: \[ \boxed{x^4+2x^2y^2+y^4} \]

Exercise 5.1 – Part 2

Q1. Find the square of the following expressions.

(f) Find the square of \( \left(\frac{3x}{2}-\frac{2y}{5}\right) \).

\[ (a-b)^2=a^2-2ab+b^2 \] \[ \left(\frac{3x}{2}-\frac{2y}{5}\right)^2 = \left(\frac{3x}{2}\right)^2 – 2\left(\frac{3x}{2}\right)\left(\frac{2y}{5}\right) + \left(\frac{2y}{5}\right)^2 \] \[ = \frac{9x^2}{4} – \frac{6xy}{5} + \frac{4y^2}{25} \]

Answer: \[ \boxed{\frac{9x^2}{4}-\frac{6xy}{5}+\frac{4y^2}{25}} \]

(g) Find the square of \( \left(2x-\frac{1}{x}\right) \).

\[ \left(2x-\frac1x\right)^2 = (2x)^2 – 2(2x)\left(\frac1x\right) + \left(\frac1x\right)^2 \] \[ = 4x^2 – 4 + \frac1{x^2} \]

Answer: \[ \boxed{4x^2-4+\frac1{x^2}} \]

(h) Find the square of \( \left(3z-\frac1z\right) \).

\[ \left(3z-\frac1z\right)^2 = (3z)^2 – 2(3z)\left(\frac1z\right) + \left(\frac1z\right)^2 \] \[ = 9z^2 – 6 + \frac1{z^2} \]

Answer: \[ \boxed{9z^2-6+\frac1{z^2}} \]

(i) Find the square of \( (a^2b-b^2a) \).

\[ (a^2b-b^2a)^2 = (a^2b)^2 – 2(a^2b)(b^2a) + (b^2a)^2 \] \[ = a^4b^2 – 2a^3b^3 + a^2b^4 \]

Answer: \[ \boxed{a^4b^2-2a^3b^3+a^2b^4} \]

(j) Find the square of \( (1.5x^2+0.3y^2) \).

\[ (a+b)^2=a^2+2ab+b^2 \] \[ (1.5x^2+0.3y^2)^2 = (1.5x^2)^2 + 2(1.5x^2)(0.3y^2) + (0.3y^2)^2 \] \[ = 2.25x^4 + 0.9x^2y^2 + 0.09y^4 \]

Answer: \[ \boxed{2.25x^4+0.9x^2y^2+0.09y^4} \]

Exercise 5.1 – Part 3

Q2. Find the value of the following.

(a) Find \( (109)^2 \).

\[ (a+b)^2=a^2+2ab+b^2 \] \[ 109^2=(100+9)^2 \] \[ =100^2+2(100)(9)+9^2 \] \[ =10000+1800+81 \] \[ =11881 \]

Answer: \[ \boxed{11881} \]

(b) Find \( (93)^2 \).

\[ 93^2=(100-7)^2 \] \[ =100^2-2(100)(7)+7^2 \] \[ =10000-1400+49 \] \[ =8649 \]

Answer: \[ \boxed{8649} \]

(c) Find \( (0.98)^2 \).

\[ (0.98)^2=(1-0.02)^2 \] \[ =1-2(0.02)+(0.02)^2 \] \[ =1-0.04+0.0004 \] \[ =0.9604 \]

Answer: \[ \boxed{0.9604} \]

(d) Find \(117\times83\).

\[ 117\times83 =(100+17)(100-17) \] \[ =100^2-17^2 \] \[ =10000-289 \] \[ =9711 \]

Answer: \[ \boxed{9711} \]

(e) Find \( (399)^2 \).

\[ 399^2=(400-1)^2 \] \[ =400^2-2(400)(1)+1^2 \] \[ =160000-800+1 \] \[ =159201 \]

Answer: \[ \boxed{159201} \]

(f) Find \(991\times1009\).

\[ 991\times1009 =(1000-9)(1000+9) \] \[ =1000^2-9^2 \] \[ =1000000-81 \] \[ =999919 \]

Answer: \[ \boxed{999919} \]

(g) Find \( (199.5)^2 \).

\[ 199.5^2=(200-0.5)^2 \] \[ =200^2-2(200)(0.5)+(0.5)^2 \] \[ =40000-200+0.25 \] \[ =39800.25 \]

Answer: \[ \boxed{39800.25} \]

(h) Find \( (0.54)^2-(0.46)^2 \).

\[ a^2-b^2=(a+b)(a-b) \] \[ =(0.54+0.46)(0.54-0.46) \] \[ =1\times0.08 \] \[ =0.08 \]

Answer: \[ \boxed{0.08} \]

Exercise 5.1 – Part 4

Q3. Evaluate.

(a) Evaluate \( (ab+cd)(ab-cd) \).

\[ (a+b)(a-b)=a^2-b^2 \] \[ (ab+cd)(ab-cd) =(ab)^2-(cd)^2 \] \[ =a^2b^2-c^2d^2 \]

Answer: \[ \boxed{a^2b^2-c^2d^2} \]

(b) Evaluate \( (x-3y)(x+3y) \).

\[ (x-3y)(x+3y) =x^2-(3y)^2 \] \[ =x^2-9y^2 \]

Answer: \[ \boxed{x^2-9y^2} \]

(c) Evaluate \( (a-0.2)(a+0.2) \).

\[ (a-0.2)(a+0.2) =a^2-(0.2)^2 \] \[ =a^2-0.04 \]

Answer: \[ \boxed{a^2-0.04} \]

(d) Evaluate \( (4.5x^2-0.9y^2)(4.5x^2+0.9y^2) \).

\[ (a-b)(a+b)=a^2-b^2 \] \[ =(4.5x^2)^2-(0.9y^2)^2 \] \[ =20.25x^4-0.81y^4 \]

Answer: \[ \boxed{20.25x^4-0.81y^4} \]

Exercise 5.1 – Part 5

Q4. Find the value of the following.

(a) Find the value of \(4x^2+49y^2+28xy\), when \(x=1,\;y=2\).

\[ 4x^2+49y^2+28xy =(2x)^2+(7y)^2+2(2x)(7y) \] \[ =(2x+7y)^2 \] Substituting \(x=1,\;y=2\), \[ =(2\times1+7\times2)^2 =(2+14)^2 =16^2 =256 \]

Answer: \[ \boxed{256} \]

(b) Find the value of \(25x^2+64y^2-80xy\), when \(x=4,\;y=3\).

\[ 25x^2+64y^2-80xy =(5x)^2+(8y)^2-2(5x)(8y) \] \[ =(5x-8y)^2 \] Substituting \(x=4,\;y=3\), \[ =(5\times4-8\times3)^2 =(20-24)^2 =(-4)^2 =16 \]

Answer: \[ \boxed{16} \]

(c) Find the value of \(4x^2+\dfrac{9}{x^2}-12\), when \(x=2\).

\[ 4x^2+\frac{9}{x^2}-12 =(2x)^2+\left(\frac3x\right)^2-2(2x)\left(\frac3x\right) \] \[ =\left(2x-\frac3x\right)^2 \] Substituting \(x=2\), \[ =\left(2\times2-\frac32\right)^2 \] \[ =\left(4-\frac32\right)^2 =\left(\frac52\right)^2 =\frac{25}{4} \] \[ =6.25 \]

Answer: \[ \boxed{\frac{25}{4}} \] or \[ \boxed{6.25} \]

Exercise 5.1 – Part 6

Q5. If \(x+\dfrac{1}{x}=3\), find the values of \(x^2+\dfrac{1}{x^2}\) and \(x^4+\dfrac{1}{x^4}\).

Given:

\[ x+\frac1x=3 \]

Step 1: Find \(x^2+\dfrac1{x^2}\).

\[ \left(x+\frac1x\right)^2 =x^2+\frac1{x^2}+2 \] \[ 3^2=x^2+\frac1{x^2}+2 \] \[ 9=x^2+\frac1{x^2}+2 \] \[ x^2+\frac1{x^2}=7 \]

Step 2: Find \(x^4+\dfrac1{x^4}\).

\[ \left(x^2+\frac1{x^2}\right)^2 =x^4+\frac1{x^4}+2 \] \[ 7^2=x^4+\frac1{x^4}+2 \] \[ 49=x^4+\frac1{x^4}+2 \] \[ x^4+\frac1{x^4}=47 \]

Answer: \[ \boxed{x^2+\frac1{x^2}=7} \] \[ \boxed{x^4+\frac1{x^4}=47} \]

Q6. If \(x+\dfrac1x=\sqrt5\), find the values of \(x^2+\dfrac1{x^2}\) and \(x^4+\dfrac1{x^4}\).

Given:

\[ x+\frac1x=\sqrt5 \]

Step 1: Find \(x^2+\dfrac1{x^2}\).

\[ \left(x+\frac1x\right)^2 =x^2+\frac1{x^2}+2 \] \[ (\sqrt5)^2=x^2+\frac1{x^2}+2 \] \[ 5=x^2+\frac1{x^2}+2 \] \[ x^2+\frac1{x^2}=3 \]

Step 2: Find \(x^4+\dfrac1{x^4}\).

\[ \left(x^2+\frac1{x^2}\right)^2 =x^4+\frac1{x^4}+2 \] \[ 3^2=x^4+\frac1{x^4}+2 \] \[ 9=x^4+\frac1{x^4}+2 \] \[ x^4+\frac1{x^4}=7 \]

Answer: \[ \boxed{x^2+\frac1{x^2}=3} \] \[ \boxed{x^4+\frac1{x^4}=7} \]

Exercise 5.1 – Part 7

Q7. If \(x^2+\dfrac{1}{x^2}=38\), find the values of \(x+\dfrac{1}{x}\) and \(x-\dfrac{1}{x}\).

Given:

\[ x^2+\frac1{x^2}=38 \]

Step 1: Find \(x+\dfrac1x\).

\[ \left(x+\frac1x\right)^2 =x^2+\frac1{x^2}+2 \] \[ =\;38+2 =40 \] \[ x+\frac1x =\sqrt{40} =2\sqrt{10} \]

Step 2: Find \(x-\dfrac1x\).

\[ \left(x-\frac1x\right)^2 =x^2+\frac1{x^2}-2 \] \[ =\;38-2 =36 \] \[ x-\frac1x =\sqrt{36} =6 \]

Answer: \[ \boxed{x+\frac1x=2\sqrt{10}} \] \[ \boxed{x-\frac1x=6} \]

Q8. If \(9x^2+25y^2=181\) and \(xy=-6\), find the value of \(3x+5y\).

Given:

\[ 9x^2+25y^2=181,\qquad xy=-6 \]

Using the identity

\[ (3x+5y)^2 =9x^2+25y^2+30xy \] \[ =181+30(-6) \] \[ =181-180 =1 \] \[ 3x+5y=\sqrt1=1 \]

Answer: \[ \boxed{3x+5y=1} \]

Q9. If \(3x-7y=10\) and \(xy=-1\), find the value of \(9x^2+49y^2\).

Given:

\[ 3x-7y=10,\qquad xy=-1 \]

Using the identity

\[ (3x-7y)^2 =9x^2+49y^2-42xy \] \[ 10^2 =9x^2+49y^2-42(-1) \] \[ 100 =9x^2+49y^2+42 \] \[ 9x^2+49y^2 =100-42 =58 \]

Answer: \[ \boxed{58} \]

Exercise 5.1 – Part 8A

Q10. Simplify.

(a) Simplify \( \left(\frac12a-3b\right)\left(3b+\frac12a\right)\left(\frac14a^2+9b^2\right) \).

\[ \left(\frac12a-3b\right) \left(\frac12a+3b\right) =\left(\frac12a\right)^2-(3b)^2 \] \[ =\frac14a^2-9b^2 \] Now, \[ \left(\frac14a^2-9b^2\right) \left(\frac14a^2+9b^2\right) \] Again using \[ (a-b)(a+b)=a^2-b^2 \] \[ =\left(\frac14a^2\right)^2-(9b^2)^2 \] \[ =\frac1{16}a^4-81b^4 \]

Answer: \[ \boxed{\frac1{16}a^4-81b^4} \]

(b) Simplify \( \left(m+\frac n5\right)^3\left(m-\frac n5\right) \).

\[ \left(m+\frac n5\right)^3 \left(m-\frac n5\right) = \left(m+\frac n5\right)^2 \left(m+\frac n5\right) \left(m-\frac n5\right) \] Using \[ (a+b)(a-b)=a^2-b^2 \] \[ = \left(m+\frac n5\right)^2 \left(m^2-\frac{n^2}{25}\right) \] This is the simplified form.

Answer: \[ \boxed{\left(m+\frac n5\right)^2 \left(m^2-\frac{n^2}{25}\right)} \]

(c) Simplify \( (x^3-3x^2-x)(x^2-3x+1) \).

Factor the first expression: \[ x^3-3x^2-x = x(x^2-3x-1) \] Therefore, \[ x(x^2-3x-1)(x^2-3x+1) \] Using \[ (a-b)(a+b)=a^2-b^2 \] where \[ a=x^2-3x,\qquad b=1 \] \[ =x\left[(x^2-3x)^2-1\right] \] Expanding, \[ =x(x^4-6x^3+9x^2-1) \] \[ =x^5-6x^4+9x^3-x \]

Answer: \[ \boxed{x^5-6x^4+9x^3-x} \]

Exercise 5.1 – Part 8B

Q10. Simplify.

(d) Simplify \( (2x^4-4x^2+1)(2x^4-4x^2-1) \).

Let \[ a=2x^4-4x^2,\qquad b=1 \] Using \[ (a+b)(a-b)=a^2-b^2 \] \[ (2x^4-4x^2+1)(2x^4-4x^2-1) =(2x^4-4x^2)^2-1 \] \[ =\left(2x^4\right)^2 -2(2x^4)(4x^2) +\left(4x^2\right)^2-1 \] \[ =4x^8-16x^6+16x^4-1 \]

Answer: \[ \boxed{4x^8-16x^6+16x^4-1} \]

(e) Simplify \(0.76\times0.76+2\times0.76\times0.24+0.24\times0.24\).

Using \[ a^2+2ab+b^2=(a+b)^2 \] where \[ a=0.76,\qquad b=0.24 \] \[ =(0.76+0.24)^2 \] \[ =1^2 =1 \]

Answer: \[ \boxed{1} \]

(f) Simplify \( \dfrac{7.83\times7.83-1.17\times1.17}{6.66} \).

Using \[ a^2-b^2=(a+b)(a-b) \] \[ =\frac{(7.83+1.17)(7.83-1.17)}{6.66} \] \[ =\frac{9\times6.66}{6.66} \] \[ =9 \]

Answer: \[ \boxed{9} \]

Exercise 5.1 – Part 10

Q11. Prove that:

\[ 2a^2+2b^2+2c^2-2ab-2bc-2ca = (a-b)^2+(b-c)^2+(c-a)^2 \] L.H.S. \[ =2a^2+2b^2+2c^2-2ab-2bc-2ca \] Now, \[ (a-b)^2=a^2-2ab+b^2 \] \[ (b-c)^2=b^2-2bc+c^2 \] \[ (c-a)^2=c^2-2ca+a^2 \] Adding, \[ (a-b)^2+(b-c)^2+(c-a)^2 \] \[ =(a^2-2ab+b^2) +(b^2-2bc+c^2) +(c^2-2ca+a^2) \] \[ =2a^2+2b^2+2c^2-2ab-2bc-2ca \] \[ =\text{L.H.S.} \] Hence, \[ \boxed{ 2a^2+2b^2+2c^2-2ab-2bc-2ca =(a-b)^2+(b-c)^2+(c-a)^2 } \] Hence Proved.

Q12. Prove that

\[ a^2+b^2+c^2-ab-bc-ca \] Using the result of Question 11, \[ (a-b)^2+(b-c)^2+(c-a)^2 = 2a^2+2b^2+2c^2-2ab-2bc-2ca \] Dividing both sides by 2, \[ a^2+b^2+c^2-ab-bc-ca = \frac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \] Since every square is non-negative, \[ (a-b)^2\ge0,\qquad (b-c)^2\ge0,\qquad (c-a)^2\ge0 \] Therefore, \[ \frac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\ge0 \] Hence, \[ \boxed{ a^2+b^2+c^2-ab-bc-ca\ge0 } \] Thus, \[ \boxed{ a^2+b^2+c^2-ab-bc-ca \text{ is always non-negative.} } \] Hence Proved.