Ncert solutions

Ncert solutions

Grade 8

Algebraic Expressions and Identities | Exercise 9.5

Question 1

Q1) Use a suitable identity to get each of the following products:

(i) \left(x+3\right)\left(x+3\right)

(ii) \left(2y+5\right)\left(2y+5\right)

(iii) \left(2a-7\right)\left(2a-7\right)

(iv) \left(3a-\frac{1}{2}\right)\left(3a-\frac{1}{2}\right)

(v) (1.1m - 0.4)(1.1m + 0.4)

(vi) \left(a^2+b^2\right)\left(-a^2+b^2\right)

(vii) (6x - 7)(6x + 7)

(viii) (-a + c)(-a + c)

(ix) \left(\frac{x}{2}+\frac{3y}{4}\right)\left(\frac{x}{2}+\frac{3y}{4}\right)

(x) (7a - 9b)(7a - 9b)

Solution:

(i) \left(x+3\right)\left(x+3\right)=\left(x+3\right)^2

= \left(x\right)^2+2\times x\times3+\left(3\right)^2

[Using identity \left(a+b\right)^2=a^2+2ab+b^2]

= x^2+6x+9

(ii) \left(2y+5\right)\left(2y+5\right)=\left(2y+5\right)^2

= \left(2y\right)^2+2\times2y\times5+\left(5\right)^2

[Using identity \left(a+b\right)^2=a^2+2ab+b^2]

= 4y^2+20y+25

(iii) \left(2a-7\right)\left(2a-7\right)=\left(2a-7\right)^2

= \left(2a\right)^2-2\times2a\times7+\left(7\right)^2

[Using identity \left(a-b\right)^2=a^2-2ab+b^2}

= 4a^2+20y+25

[Using identity \left(a-b\right)^2=a^2-2ab+b^2]

= 4a^2-28a+49

(iv) \left(3a-\frac{1}{2}\right)\left(3a-\frac{1}{2}\right)=\left(3a-\frac{1}{2}\right)^2

= \left(3a\right)^2-2\times3a\times\frac{1}{2}+\left(\frac{1}{2}\right)^2

[Using identity \left(a-b\right)^2=a^2-2ab+b^2

= 9a^2-3a+\frac{1}{4}

(v) \left(1.1m-0.4\right)\left(1.1m+0.4\right)=\left(1.1m\right)^2-\left(0.4\right)^2

[Using identity \left(a-b\right)\left(a+b\right)=a^2-b^2]

= 1.21m^3-0.16

(vi) \left(a^2+b^2\right)\left(-a^2+b^2\right)=a^2-b^2]

= \left(b^2\right)^{^2}-\left(a^2\right)^2

[Using identity \left(a-b\right)\left(a+b\right)=a^2-b^2]

= b^4-a^4

(vii) \left(6x-7\right)\left(6x+7\right)=\left(6x\right)^2-\left(7\right)^2

[Using identity \left(a-b\right)\left(a+b\right)=a^2-b^2

= 36x^2-49

(viii) \left(-a+c\right)\left(-a+c\right)

\left(c-a\right)\left(c-a\right)=\left(c-a\right)^2

= \left(c\right)^2-2\times c\times c+\left(a\right)^2

[Using identity \left(a-b\right)^2=a^2-2ab+b^2]

= c^2-2\times c\times a+\left(a\right)^2]

= c^2-2ca+a^2

(ix) \left(\frac{x}{2}+\frac{3y}{4}\right)\left(\frac{x}{2}+\frac{3y}{4}\right)=\left(\frac{x}{2}+\frac{3y}{4}\right)^2

= \left(\frac{x}{2}\right)^2+2\times\frac{x}{2}\times\frac{3y}{4}+\left(\frac{3y}{4}\right)^2

[Using identity \left(a+b\right)^2=a^2+2ab+b^2

= \frac{x^2}{4}+\frac{3}{4}xy+\frac{9}{16}y^2

(x) \left(7a-9b\right)\left(7a-9b\right)=\left(7a-9b\right)^2

= \left(7a\right)^2-2\times7a\times9b+\left(9b\right)^2

[Using identity \left(a-b\right)^2=a^2-2ab+b^2]

= 49a^2-126ab+81b^2

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Question 1

Q1) Use a suitable identity to get each of the following products:

(i) \left(x+3\right)\left(x+3\right)

(ii) \left(2y+5\right)\left(2y+5\right)

(iii) \left(2a-7\right)\left(2a-7\right)

(iv) \left(3a-\frac{1}{2}\right)\left(3a-\frac{1}{2}\right)

(v) (1.1m - 0.4)(1.1m + 0.4)

(vi) \left(a^2+b^2\right)\left(-a^2+b^2\right)

(vii) (6x - 7)(6x + 7)

(viii) (-a + c)(-a + c)

(ix) \left(\frac{x}{2}+\frac{3y}{4}\right)\left(\frac{x}{2}+\frac{3y}{4}\right)

(x) (7a - 9b)(7a - 9b)

Solution:

(i) \left(x+3\right)\left(x+3\right)=\left(x+3\right)^2

= \left(x\right)^2+2\times x\times3+\left(3\right)^2

[Using identity \left(a+b\right)^2=a^2+2ab+b^2]

= x^2+6x+9

(ii) \left(2y+5\right)\left(2y+5\right)=\left(2y+5\right)^2

= \left(2y\right)^2+2\times2y\times5+\left(5\right)^2

[Using identity \left(a+b\right)^2=a^2+2ab+b^2]

= 4y^2+20y+25

(iii) \left(2a-7\right)\left(2a-7\right)=\left(2a-7\right)^2

= \left(2a\right)^2-2\times2a\times7+\left(7\right)^2

[Using identity \left(a-b\right)^2=a^2-2ab+b^2}

= 4a^2+20y+25

[Using identity \left(a-b\right)^2=a^2-2ab+b^2]

= 4a^2-28a+49

(iv) \left(3a-\frac{1}{2}\right)\left(3a-\frac{1}{2}\right)=\left(3a-\frac{1}{2}\right)^2

= \left(3a\right)^2-2\times3a\times\frac{1}{2}+\left(\frac{1}{2}\right)^2

[Using identity \left(a-b\right)^2=a^2-2ab+b^2

= 9a^2-3a+\frac{1}{4}

(v) \left(1.1m-0.4\right)\left(1.1m+0.4\right)=\left(1.1m\right)^2-\left(0.4\right)^2

[Using identity \left(a-b\right)\left(a+b\right)=a^2-b^2]

= 1.21m^3-0.16

(vi) \left(a^2+b^2\right)\left(-a^2+b^2\right)=a^2-b^2]

= \left(b^2\right)^{^2}-\left(a^2\right)^2

[Using identity \left(a-b\right)\left(a+b\right)=a^2-b^2]

= b^4-a^4

(vii) \left(6x-7\right)\left(6x+7\right)=\left(6x\right)^2-\left(7\right)^2

[Using identity \left(a-b\right)\left(a+b\right)=a^2-b^2

= 36x^2-49

(viii) \left(-a+c\right)\left(-a+c\right)

\left(c-a\right)\left(c-a\right)=\left(c-a\right)^2

= \left(c\right)^2-2\times c\times c+\left(a\right)^2

[Using identity \left(a-b\right)^2=a^2-2ab+b^2]

= c^2-2\times c\times a+\left(a\right)^2]

= c^2-2ca+a^2

(ix) \left(\frac{x}{2}+\frac{3y}{4}\right)\left(\frac{x}{2}+\frac{3y}{4}\right)=\left(\frac{x}{2}+\frac{3y}{4}\right)^2

= \left(\frac{x}{2}\right)^2+2\times\frac{x}{2}\times\frac{3y}{4}+\left(\frac{3y}{4}\right)^2

[Using identity \left(a+b\right)^2=a^2+2ab+b^2

= \frac{x^2}{4}+\frac{3}{4}xy+\frac{9}{16}y^2

(x) \left(7a-9b\right)\left(7a-9b\right)=\left(7a-9b\right)^2

= \left(7a\right)^2-2\times7a\times9b+\left(9b\right)^2

[Using identity \left(a-b\right)^2=a^2-2ab+b^2]

= 49a^2-126ab+81b^2

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