Finding square root using prime factorization method part 2

  Q.1 Find the square root of 625 using the prime factorization method  Solution:  625=5×5×5×5  Make pairs 625=5×5×5×5  √625=√5×5×5×5  √625=5×5 √625=25  Q.2 Find the square root of 144  using the prime factorization method  Solution:  144=2×2×2×2×3×3 Make pairs  144=2×2×2×2×3×3 √144=√2×2×2×2×3×3 √144=2×3×2 √144=12 Q.3 Find the square root of 400 using the prime factorization method  Solution:  400=2×2×2×2×5×5 … Read more

Finding Square root using prime factorization method

Procedure to find square root using prime factorization method  Write the prime factors of the given number  Make pairs of the equal factors  Write one factor corresponding to each pair  Multiply the obtained factors Q.1 Find the square root of 256 using the prime factorization method. Solution: 256=2×2×2×2×2×2×2×2 Make pairs  256=2×2×2×2×2×2×2×2 √256=√2×2×2×2×2×2×2×2 √256=2×2×2×2 √256=16  Q.2 … Read more

Worksheet on algebraic identities with answer

Here are some questions solution of algebraic identities 

Q.1 Write the following in the expanded form 

(i) (3x+4y)²

(ii) (2x-3y)(2x+3y) 

(iii) (y-6)(y-9) 

Q.2 Using algebraic identity evaluate : 

(i) (98)²

(ii) 48×52

(iii) 101 ×103 

(iv) 153×153 -47×47 

Q.3 If x-\frac{1}{x}=5 ,find the value of the 

(i) x²+\frac{1}{x²}

(ii) x+\frac{1}{x}

Q.4 Find z ,if 9z=79²-61²

Q.5 Show that 

(i) (2x-7y)²+28xy=4x²+49y²

(ii) (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 

Solution: 

Problem: 01

Write the following in the expanded form 

(i) (3x+4y)²

(ii) (2x-3y)(2x+3y) 

(iii) (y-6)(y-9) 

Solution:(i) 

Using the identity :  (x+y)²=x²+y²+2xy 

(3x+4y)²=(3x)²+(4y)²+2.(3x).(4y) 

(3x+4y)²=9x²+16y²+24xy 

Solution:(ii) 

Using the identity : (x-y)(x+y)=x²-y²

(2x-3y)(2x+3y) =(2x)²-(3y)²

(2x-3y)(2x+3y) =4x²-9y²

Solution:(iii) 

Using the identity : (x+a)(x+b)=x²+(a+b)x+ab 

(y-6)(y-9) 

Here a=-6 and b=-9 

(y-6)(y-9) =y²+{-6+(-9)}y+(-6)(-9) 

(y-6)(y-9) =y²+{-6-9}y+54 

(y-6)(y-9) =y²-15y+54 

Problem: 02

Using algebraic identity evaluate : 

(i) (98)²

(ii) 48×52

(iii) 101 ×103 

(iv) 153×153 -47×47 

Solution:(i) 

(98)² can be written as (100-2)² which is of form (a-b)²

Using the identity : (a-b)²=a²-2ab+b²

(100-2)²=(100)²-2.100.2.+(2)²

(100-2)²=10000-400+4 

(100-2)²=9604 

Solution(ii) 

48×52 can be written as (50-2)×(50+2) which is of form (a-b)(a+b) 

Using the identity : (a-b)(a+b)=a²-b²

(50-2)×(50+2)=(50)²-(2)²

(50-2)×(50+2)=2500-4 

(50-2)×(50+2)=2496 

Solution(iii)

101 ×103 can be written as (100+1)(100+3) which is of form (x+a)(x+b) 

Using the identity : (x+a)(x+b) =x²+(a+b)x+ab 

 We have a=1 and b=3 

(100+1)(100+3)=(100)²+(1+3)(100)+1.3

(100+1)(100+3)=10000+(4)(100)+3 

(100+1)(100+3)=10000+400+3 

(100+1)(100+3)=10403 

Solution(iv) 

153×153 -47×47  is of form a.a-b.b=a²-b² 

Using identity : a²-b²=(a+b)(a-b) 

153×153 -47×47 =(153)²-(47)²

(153)²-(47)²=(153+47)(153-47) 

(153)²-(47)²=106×200

(153)²-(47)²=21200 

Problem:03

If x-\frac{1}{x}=5 ,find the value of the 

(i) x²+\frac{1}{x²}

(ii) x+\frac{1}{x}

Solution(i) 

We have  , x-\frac{1}{x}=5

Squaring on both sides 

(x-\frac{1}{x})²=(5)²

Using identity : (x-y)²=x²-2xy+y² 

(x)²-2.x.\frac{1}{x}+(\frac{1}{x²})=25 

x²-2.+(\frac{1}{x²})=25 

x²+\frac{1}{x²}-2=25 

x²+\frac{1}{x²}=27 

Thus ,value of x²+\frac{1}{x²}=27                  ………………(i) 

Solution(ii) 

Find the value of (x+\frac{1}{x}

(x+\frac{1}{x})²=x²+2.x.\frac{1}{x}+\frac{1}{x²}

(x+\frac{1}{x})²=x²+2+\frac{1}{x²}                  ………………….(ii) 

We have value of x²+\frac{1}{x²}=27  from eq(i) 

Substitute in eq(ii) 

(x+\frac{1}{x})²=27+2 

(x+\frac{1}{x})²=29 

x+\frac{1}{x}=√29 

Problem: 04 Find z ,if 9z=79²-61²

Solution: 

We have , 9z=79²-61²

Using the identity :a²-b²=(a+b)(a-b) 

So, 79²-61²=(79+61)(79-61) =(140)(18) 

9z= 140.18 

z=\frac{140.18}{9}

z= 280 

Problem: 05

Show that 

(i) (2x-7y)²+28xy=4x²+49y²

(ii) (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 

Solution(i) :

(2x-7y)²+28xy 

Exapand Using identity : (x-y)²=x²-2xy+y²

=(2x)²-2.(2x).(7y)+(7y)²+28xy 

= 4x²-28xy+49y²+28xy 

=4x²+49y²+0 

=4x²+49y²

Hence proved (2x-7y)²+28xy=4x²+49y²

Solution(ii) 

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 

Using identity : (x-y)(x+y)=x²-y²

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)

(a-b)(a+b)=a²-b²,(b-c)(b+c)=b²-c² and (c-a)(c+a)=c²-a²

=a²-b²+b²-c²+c²-a²

=0

Hence proved  ,(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 

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Worksheet on Multiplication of binomials

Q.1 Simplify  (i) (a²+b²)(a+b)  (ii) (3x²-y²)(10x²-3y²) (iii) 3x(-6x+11y)-(5x+12y)(2x+9y)  (iv) 7x(2x-3)-9x(2x²)-5x(2x+3)  Q.2 Simplify this  (7x-9y²)(2x²-3y)-(5x+2y)(9x-10y)-2x(3x+5y)  Q.3 Find the following products  (i) (10a-11b)(5a+1)  (ii) (1 – x3) × (x2 + 2) Answer Problem: 01 Simplify  (i) (a²+b²)(a+b)  (ii) (3x²-y²)(10x²-3y²) (iii) 3x(-6x+11y)-(5x+12y)(2x+9y)  (iv) 7x(2x-3)-9x(2x²)-5x(2x+3)  Solution:(i)  (a²+b²)(a+b) =a²(a+b)+b²(a+b)  (a²+b²)(a+b) =a³+a²b+b²a+b³ (a²+b²)(a+b) =a³+b³+ab(a+b)  Solution :(ii)  (3x²-y²)(10x²-3y²)=3x²(10x²-3y²)-y²(10x²-3y²) (3x²-y²)(10x²-3y²)=30-9x²y²-10x²y²+3 (3x²-y²)(10x²-3y²)=30+3-19x²y² Solution:(iii)  3x(-6x+11y)-(5x+12y)(2x+9y) =(-18x²+33xy)-{5x(2x+9y)+12x(2x+9y)} … Read more

Worksheet on multiplication of binomial with a monomial

Q.1 Multiply (i) -9xy by -2x²y+3x  (ii) 2x³y by (-3xy+xy²)  Q.2 Multiply -3x²y by 7x³y -5xy² .Find the value of the product by taking x=- 2 and  y=-1  Q.3 Find the product of the following and find the value of the product by taking x= -3 and y=2  (i) 3x and y+13x  (ii) 5xy and … Read more

Worksheet on addition of polynomial with solution

Q.1 Add (i) 41x²-6xy+12y²  and 11x²+2xy-7y²  (ii) 9x²-10xy-5xy² ,3x²+15xy-2xy²  and -2x²+3xy²-2xy  Q.2 What should be added to 12x²+3x-2 to obtain 3x² +9x+5 ?  Q.3 Subtract the sum of 4x²-3xy+y² and -2xy +9x²-5y² from -8x²+5xy  Q.4 Find the sum of 5x²-7xy+4y²-3x, 4x²+2xy-y², and x²+5xy-2y² +3x-y  Q.5 What must be added to x² +4x -6 to get … Read more