# NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.5 in Chapter 9 - Algebraic Expressions and Identities

Question 1 Exercise 9.5

Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3)

(ii) (2y + 5) (2y + 5)

(iii) (2a - 7) (2a - 7)

(iv) (3a - 1/2)(3a - 1/2)

(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{1}{2} x+\frac{3}{4} y\right)\left(\frac{1}{2} x+\frac{3}{4} y\right)

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

\begin{aligned} &\text { i) } \quad(x+3)(x+3)=(x+3)^{2}\\ &\begin{array}{l} =x^{2}+6 x+9 \\ \text { Using }(a+b)^{2}=a^{2}+b^{2}+2 a b \end{array} \end{aligned}

\begin{aligned} &\text { ii) }\\ &\begin{array}{c} (2 y+5)(2 y+5)=(2 y+5)^{2} \\ =4 y^{2}+20 y+25 \\ \text { Using }(a+b)^{2}=a^{2}+b^{2}+2 a b \end{array} \end{aligned}

\begin{aligned} &\text { iii) } \quad(2 a-7)(2 a-7)=(2 a-7)^{2}\\ &\begin{array}{l} =4 a^{2}-28 a+49 \\ \operatorname{Using}(a-b)^{2}=a^{2}+b^{2}-2 a b \end{array} \end{aligned}

\begin{aligned} &\text { iv) } \quad(3 a-1 / 2)(3 a-1 / 2)=(3 a-1 / 2)^{2}\\ &=(3 a-1 / 2)(3 a-1 / 2)=9 a^{2}-3 a+(1 / 4) \end{aligned}

\text { Using }(a-b)^{2}=a^{2}+b^{2}-2 a b

\begin{aligned} &\mathrm{v})\\ &(1.1 \mathrm{m}-0.4)(1.1 \mathrm{m}+0.4)\\ &\begin{array}{c} =1.21 \mathrm{m}^{2}-0.16 \\ \operatorname{Using}(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})=\mathrm{a}^{2}-\mathrm{b}^{2} \end{array} \end{aligned}

\begin{aligned} &\text { vi) }\left(a^{2}+b^{2}\right)\left(-a^{2}+b^{2}\right)\\ &=\left(b^{2}+a^{2}\right)\left(b^{2}-a^{2}\right)\\ &=-a^{4}+b^{4}\\ &\operatorname{Using}(a-b)(a+b)=a^{2}-b^{2} \end{aligned}

\begin{aligned} &\text { vii) }\\ &\begin{array}{c} (6 x-7)(6 x+7) \\ =36 x^{2}-49 \\ \operatorname{Using}(a-b)(a+b)=a^{2}-b^{2} \end{array} \end{aligned}

\begin{aligned} &\text { viii) }(-a+c)(-a+c)=(-a+c)^{2}\\ &=c^{2}+a^{2}-2 a c\\ &\operatorname{Using}(a-b)^{2}=a^{2}+b^{2}-2 a b \end{aligned}

\begin{aligned} &\text { ix ) }\left(\frac{1}{2} x+\frac{3}{4} y\right)\left(\frac{1}{2} x+\frac{3}{4} y\right)=\left(\frac{1}{2} x+\frac{3}{4} y\right)^{2}\\ &=\left(x^{2} / 4\right)+\left(9 y^{2} / 16\right)+(3 x y / 4) \end{aligned}

\text { Using }(a+b)^{2}=a^{2}+b^{2}+2 a b

\begin{aligned} &\text { x) }(7 a-9 b)(7 a-9 b)=(7 a-9 b)^{2}\\ &=49 a^{2}-126 a b+81 b^{2} \end{aligned}

\text { Using }(a-b)^{2}=a^{2}+b^{2}-2 a b

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