RD Sharma solution class 8 chapter 6 Algebraic Expressions and Identity Exercise 6.6

Exercise 6.6

Page-6.43

Question 1:

Write the following squares of binomials as trinomials:
(i) (x + 2)²
(ii) (8a + 3b)²
(iii) (2m + 1)²
(iv) ${\left(9a+\frac{1}{6}\right)}^2$
(v) ${\left(x+\frac{x^2}{2}\right)}^2$
(vi) $\left(\frac{x}{4}-\frac{y}{3}\right)$
(vii) ${\left(3x-\frac{1}{3x}\right)}^2$
(viii) ${\left(\frac{x}{y}-\frac{y}{x}\right)}^2$
(ix) ${\left(\frac{3a}{2}-\frac{5b}{4}\right)}^2$
(x) (a²b − bc²)²
(xi) ${\left(\frac{2a}{3b}+\frac{2b}{3a}\right)}^2$
(xii) (x² − ay)²

Answer 1:

 We will use the identities ${\left(a+b\right)}^2=a^2+2ab+b^2 \text{and} {\left(a-b\right)}^2=a^2-2ab+b^2$ to convert the squares of binomials as trinomials.

$$\begin{gathered} \left(\mathrm{i}\right) {\left(x+2\right)}^2 \\ =x^2+2\times x\times 2+b^2 \\ =x^2+4x+b^2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) {\left(8a+3b\right)}^2 \\ ={\left(8a\right)}^2+2\left(8a\right)\left(3b\right)+{\left(6b\right)}^2 \\ =64a^2+48ab+36b^2 \end{gathered}$$

<sup>$$\begin{gathered} \left(\mathrm{iii}\right) {\left(2m+1\right)}^2 \\ ={\left(2m\right)}^2+2\left(2m\right)\left(1\right)+1^2 \\ =4m^2+4m+1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) {\left(9a+\frac{1}{6}\right)}^2 \\ ={\left(9a\right)}^2+2\left(9a\right)\left(\frac{1}{6}\right)+{\left(\frac{1}{6}\right)}^2 \\ =81a^2+3a+\frac{1}{36} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right) {\left(x+\frac{x^2}{2}\right)}^2 \\ =x^2+2x\left(\frac{x^2}{2}\right)+{\left(\frac{x^2}{2}\right)}^2 \\ =x^2+x^3+\frac{x^4}{4} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right) {\left(\frac{x}{4}-\frac{y}{3}\right)}^2 \\ ={\left(\frac{x}{4}\right)}^2-2\left(\frac{x}{4}\right)\left(\frac{y}{3}\right)+{\left(\frac{y}{3}\right)}^2 \\ =\frac{x^2}{16}-\frac{1}{6}xy+\frac{y^2}{9} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{vii}\right) {\left(3x-\frac{1}{3x}\right)}^2 \\ ={\left(3x\right)}^2-2\left(3x\right)\left(\frac{1}{3x}\right)+{\left(\frac{1}{3x}\right)}^2 \\ =9x^2-2+\frac{1}{9x^2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right) {\left(\frac{x}{y}-\frac{y}{x}\right)}^2 \\ ={\left(\frac{x}{y}\right)}^2-2\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)+{\left(\frac{y}{x}\right)}^2 \\ =\frac{x^2}{y^2}-2+\frac{y^2}{x^2} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ix}\right) {\left(\frac{3a}{2}-\frac{5b}{4}\right)}^2 \\ ={\left(\frac{3a}{2}\right)}^2-2\left(\frac{3a}{2}\right)\left(\frac{5b}{4}\right)+{\left(\frac{5b}{4}\right)}^2 \\ =\frac{9a^2}{4}-\frac{15ab}{4}+\frac{25b^2}{16} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{x}\right) {\left(a^{2}b-b{c}^2\right)}^2 \\ ={\left(a^{2}b\right)}^2-2\left(a^{2}b\right)\left(b{c}^2\right)+{\left(b{c}^2\right)}^2 \\ =a^4{b}^2-2a^2{b}^2{c}^2+b^2{c}^4 \end{gathered}$$</sup>



$$\begin{gathered} \left(\mathrm{xi}\right) {\left(\frac{2a}{3b}+\frac{2b}{3a}\right)}^2 \\ ={\left({\displaystyle \frac{2a}{3b}}\right)}^2+2\left({\displaystyle \frac{2a}{3b}}\right)\left(\frac{2b}{3a}\right)+{\left(\frac{2b}{3a}\right)}^2 \\ =\frac{4a^2}{9b^2}+\frac{8}{9}+\frac{4b^2}{9a^2} \end{gathered}$$



$$\begin{gathered} \left(\mathrm{xii}\right) {\left(x^2-ay\right)}^2 \\ ={\left(x^2\right)}^2-2x^2\left(ay\right)+{\left(ay\right)}^2 \\ =x^4-2x^{2}ay+a^2{y}^2 \end{gathered}$$

Question 2:

Find the product of the following binomials:
(i) (2x + y)(2x + y)
(ii) (a + 2b)(a − 2b)
(iii) (a² + bc)(a² − bc)
(iv) $\left(\frac{4x}{5}-\frac{3y}{4}\right)\left(\frac{4x}{5}+\frac{3y}{4}\right)$
(v) $\left(2x+\frac{3}{y}\right)\left(2x-\frac{3}{y}\right)$
(vi) (2a³ + b³)(2a³ − b³)
(vii) $\left(x^4+\frac{2}{x^2}\right)\left(x^4-\frac{2}{x^2}\right)$
(viii) $\left(x^3+\frac{1}{x^3}\right)\left(x^3-\frac{1}{x^3}\right)$

Answer 2:

(i) We will use the identity ${\left(a+b\right)}^2=a^2+2ab+b^2$  in the given expression to find the product.
$$\begin{gathered} \left(2x+y\right)\left(2x+y\right) \\ ={\left(2x+y\right)}^2 \\ ={\left(2x\right)}^2+2\left(2x\right)\left(y\right)+y^2 \\ =4x^2+4xy+y^2 \end{gathered}$$

(ii) We will use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ in the given expression to find the product.
$$\begin{gathered} \left(a+2b\right)\left(a-2b\right) \\ =a^2-{\left(2b\right)}^2 \\ =a^2-4b^2 \end{gathered}$$

(iii) We will use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ in the given expression to find the product.
$$\begin{gathered} \left(a^2+bc\right)\left(a^2-bc\right) \\ ={\left(a^2\right)}^2-{\left(bc\right)}^2 \\ =a^4-b^2{c}^2 \end{gathered}$$

(iv)We will use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ in the given expression to find the product.
$$\begin{gathered} \left(\frac{4x}{5}-\frac{3y}{4}\right)\left(\frac{4x}{5}+\frac{3y}{4}\right) \\ ={\left(\frac{4x}{5}\right)}^2-{\left(\frac{3y}{4}\right)}^2 \\ =\frac{16x^2}{25}-\frac{9y^2}{16} \end{gathered}$$

(v) We will use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ in the given expression to find the product.
$$\begin{gathered} \left(2x+\frac{3}{y}\right)\left(2x-\frac{3}{y}\right) \\ ={\left(2x\right)}^2-{\left(\frac{3}{y}\right)}^2 \\ =4x^2-\frac{9}{y^2} \end{gathered}$$

(vi) We will use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ in the given expression to find the product.
$$\begin{gathered} \left(2a^3+b^3\right)\left(2a^3-b^3\right) \\ ={\left(2a^3\right)}^2-{\left(b^3\right)}^2 \\ =4a^6-b^6 \end{gathered}$$

(vii) We will use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ in the given expression to find the product.
$$\begin{gathered} \left(x^4+\frac{2}{x^2}\right)\left(x^4-\frac{2}{x^2}\right) \\ ={\left(x^4\right)}^2-{\left(\frac{2}{x^2}\right)}^2 \\ =x^8-\frac{4}{x^4} \end{gathered}$$

(viii) We will use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ in the given expression to find the product.

$$\begin{gathered} \left(x^3+\frac{1}{x^3}\right)\left(x^3-\frac{1}{x^3}\right) \\ ={\left(x^3\right)}^2-{\left(\frac{1}{x^3}\right)}^2 \\ =x^6-\frac{1}{x^6} \end{gathered}$$

Question 3:

Using the formula for squaring a binomial, evaluate the following:
(i) (102)²
(ii) (99)²
(iii) (1001)²
(iv) (999)²
(v) (703)²

Answer 3:

(i) Here, we will use the identity ${\left(a+b\right)}^2=a^2+2ab+b^2$
$$\begin{gathered} {\left(102\right)}^2 \\ ={\left(100+2\right)}^2 \\ ={\left(100\right)}^2+2\times 100\times 2+2^2 \\ =10000+400+4 \\ =10404 \end{gathered}$$

(ii) Here, we will use the identity ${\left(a-b\right)}^2=a^2-2ab+b^2$
$$\begin{gathered} {\left(99\right)}^2 \\ ={\left(100-1\right)}^2 \\ ={\left(100\right)}^2-2\times 100\times 1+1^2 \\ =10000-200+1 \\ =9801 \end{gathered}$$

(iii) Here, we will use the identity ${\left(a+b\right)}^2=a^2+2ab+b^2$
$$\begin{gathered} {\left(1001\right)}^2 \\ ={\left(1000+1\right)}^2 \\ ={\left(1000\right)}^2+2\times 1000\times 1+1^2 \\ =1000000+2000+1 \\ =1002001 \end{gathered}$$

(iv) Here, we will use the identity ${\left(a-b\right)}^2=a^2-2ab+b^2$
$$\begin{gathered} {\left(999\right)}^2 \\ ={\left(1000-1\right)}^2 \\ ={\left(1000\right)}^2-2\times 1000\times 1+1^2 \\ =1000000-2000+1 \\ =998001 \end{gathered}$$

(v) Here, we will use the identity ${\left(a+b\right)}^2=a^2+2ab+b^2$
$$\begin{gathered} {\left(703\right)}^2 \\ ={\left(700+3\right)}^2 \\ ={\left(700\right)}^2+2\times 700\times 3+3^2 \\ =490000+4200+9 \\ =494209 \end{gathered}$$

Question 4:

Simplify the following using the formula: (a − b)(a + b) = a² − b²:
(i) (82)² − (18)²
(ii) (467)² − (33)²
(iii) (79)² − (69)²
(iv) 197 × 203
(v) 113 × 87
(vi) 95 × 105
(vii) 1.8 × 2.2
(viii) 9.8 × 10.2

Answer 4:

Here, we will use the identity $(a-b)(a+b)=a^{2 }-b^2$

(i) Let us consider the following expression:

$$\begin{gathered} {\left(82\right)}^2-{\left(18\right)}^2 \\ =\left(82+18\right)\left(82-18\right) \\ =100\times 64 \\ =6400 \end{gathered}$$

(ii) Let us consider the following expression:

$$\begin{gathered} {\left(467\right)}^2-{\left(33\right)}^2 \\ =\left(467+33\right)\left(467-33\right) \\ =500\times 434 \\ =217000 \end{gathered}$$

(iii) Let us consider the following expression:

$$\begin{gathered} {\left(79\right)}^2-{\left(69\right)}^2 \\ =\left(79+69\right)\left(79-69\right) \\ =148\times 10 \\ =1480 \end{gathered}$$

(iv) Let us consider the following product:

$197\times 203$

$\because \frac{197+203}{2}=\frac{400}{2}=200$; therefore, we will write the above product as:
$$\begin{gathered} 197\times 203 \\ =\left(200-3\right)\left(200+3\right) \\ ={\left(200\right)}^2-{\left(3\right)}^2 \\ =40000-9 \\ =39991 \end{gathered}$$

Thus, the answer is $39991$.

(v) Let us consider the following product:

$113\times 87$

$\because \frac{113+87}{2}=\frac{200}{2}=100$; therefore, we will write the above product as:
$$\begin{gathered} 113\times 87 \\ =\left(100+13\right)\left(100-13\right) \\ ={\left(100\right)}^2-{\left(13\right)}^2 \\ =10000-169 \\ =9831 \end{gathered}$$

Thus, the answer is 9831.

(vi) Let us consider the following product:

$95\times 105$

$\because \frac{95+105}{2}=\frac{200}{2}=100$; therefore, we will write the above product as:
$$\begin{gathered} 95\times 105 \\ =\left(100+5\right)\left(100-5\right) \\ ={\left(100\right)}^2-{\left(5\right)}^2 \\ =10000-25 \\ =9975 \end{gathered}$$

Thus, the answer is 9975.

(vii) Let us consider the following product:

$1.8\times 2.2$

$\because \frac{1.8+2.2}{2}=\frac{4}{2}=2$; therefore, we will write the above product as:
$$\begin{gathered} 1.8\times 2.2 \\ =\left(2-0.2\right)\left(2+0.2\right) \\ ={\left(2\right)}^2-{\left(0.2\right)}^2 \\ =4-0.04 \\ =3.96 \end{gathered}$$

Thus, the answer is 3.96.

(viii) Let us consider the following product:

$9.8\times 10.2$

$\because \frac{9.8+10.2}{2}=\frac{20}{2}=10$; therefore, we will write the above product as:
$$\begin{gathered} 9.8\times 10.2 \\ =\left(10-0.2\right)\left(10+0.2\right) \\ ={\left(10\right)}^2-{\left(0.2\right)}^2 \\ =100-0.04 \\ =99.96 \end{gathered}$$

Thus, the answer is 99.96.

Question 5:

Simplify the following using the identities:
(i) $\frac{{58}^2-{42}^2}{16}$
(ii) 178 × 178 − 22 × 22
(iii) $\frac{198\times 198-102\times 102}{96}$
(iv) 1.73 × 1.73 − 0.27 × 0.27
(v) $\frac{8.63\times 8.63-1.37\times 1.37}{0.726}$

Answer 5:

(i) Let us consider the following expression:

$\frac{{58}^2-{42}^2}{16}$
Using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$, we get:

$\frac{{58}^2-{42}^2}{16}=\frac{\left(58+42\right)\left(58-42\right)}{16}$
$$\begin{gathered} \Rightarrow \frac{{58}^2-{42}^2}{16}=\frac{100\times 16}{16} \\ \Rightarrow \frac{{58}^2-{42}^2}{16}=100 \end{gathered}$$

Thus, the answer is 100.

(ii) Let us consider the following expression:

$178\times 178-22\times 22$
Using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$, we get:

$178\times 178-22\times 22={178}^2-{22}^2=\left(178+22\right)\left(178-22\right)=200\times 156=31200$
Thus, the answer is 31200.

(iii) Let us consider the following expression:

$\frac{198\times 198-102\times 102}{96}=\frac{{198}^2-{102}^2}{96}$
Using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$, we get:

$\frac{198\times 198-102\times 102}{96}=\frac{{198}^2-{102}^2}{96}=\frac{\left(198+102\right)\left(198-102\right)}{96}$
$$\begin{gathered} \Rightarrow \frac{198\times 198-102\times 102}{96}=\frac{\left(198+102\right)\left(198-102\right)}{96} \\ \Rightarrow \frac{198\times 198-102\times 102}{96}=\frac{300\times 96}{96} \\ \Rightarrow \frac{198\times 198-102\times 102}{96}=300 \end{gathered}$$

Thus, the answer is 300.

(iv) Let us consider the following expression:

$1.73\times 1.73-0.27\times 0.27$
Using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$, we get:

$1.73\times 1.73-0.27\times 0.27=1.{73}^2-0.{27}^2=\left(1.73+0.27\right)\left(1.73-0.27\right)=2\times 1.46=2.92$
Thus, the answer is 2.92.

(v) Let us consider the following expression:

$\frac{8.63\times 8.63-1.37\times 1.37}{0.726}=\frac{8.{63}^2-1.{37}^2}{0.726}$
Using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$, we get:

$\frac{8.63\times 8.63-1.37\times 1.37}{0.726}=\frac{8.{63}^2-1.{37}^2}{0.726}=\frac{\left(8.63+1.37\right)\left(8.63-1.37\right)}{0.726}$
$$\begin{gathered} \Rightarrow \frac{8.63\times 8.63-1.37\times 1.37}{0.726}=\frac{\left(8.63+1.37\right)\left(8.63-1.37\right)}{0.726} \\ \Rightarrow \frac{8.63\times 8.63-1.37\times 1.37}{0.726}=\frac{\left(8.63+1.37\right)\left(8.63-1.37\right)}{0.726} \\ \Rightarrow \frac{8.63\times 8.63-1.37\times 1.37}{0.726}=\frac{10\times 7.26}{0.726} \\ \Rightarrow \frac{8.63\times 8.63-1.37\times 1.37}{0.726}=\frac{10\times {\cancel{7.26}}^{10}}{\cancel{0.726}} \\ \Rightarrow \frac{8.63\times 8.63-1.37\times 1.37}{0.726}=100 \end{gathered}$$

Thus, the answer is 100.

Question 6:

Find the value of x, if:
(i) 4x = (52)² − (48)²
(ii) 14x = (47)² − (33)²
(iii) 5x = (50)² − (40)²

Answer 6:

(i) Let us consider the following equation:

$4x={\left(52\right)}^2-{\left(48\right)}^2$
Using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$, we get:
$$\begin{gathered} 4x={\left(52\right)}^2-{\left(48\right)}^2 \\ 4x=\left(52+48\right)\left(52-48\right) \\ 4x=100\times 4=400 \end{gathered}$$
$\Rightarrow 4x=400$
$\Rightarrow x=100$        (Dividing both sides by 4)

(ii) Let us consider the following equation:

$14x={\left(47\right)}^2-{\left(33\right)}^2$
Using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$, we get:
$$\begin{gathered} 14x={\left(47\right)}^2-{\left(33\right)}^2 \\ 14x=\left(47+33\right)\left(47-33\right) \\ 14x=80\times 14=1120 \end{gathered}$$
$\Rightarrow 14x=1120$
$\Rightarrow x=80$        (Dividing both sides by 14)

(iii) Let us consider the following equation:

$5x={\left(50\right)}^2-{\left(40\right)}^2$
Using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$, we get:
$$\begin{gathered} 5x={\left(50\right)}^2-{\left(40\right)}^2 \\ 5x=\left(50+40\right)\left(50-40\right) \\ 5x=90\times 10=900 \end{gathered}$$
$\Rightarrow 5x=900$
$\Rightarrow x=180$        (Dividing both sides by 5)

Question 7:

If $x+\frac{1}{x}=20,$ find the value of $x^2+\frac{1}{x^2}.$

Answer 7:

Let us consider the following equation:

$x+\frac{1}{x}=20$
Squaring both sides, we get:

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2={\left(20\right)}^2=400 \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2=400 \\ \Rightarrow x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=400 \left[\right(a+b{)}^2=a^{2 }+b^2 +2ab] \\ \Rightarrow x^2+2+\frac{1}{x^2}=400 \end{gathered}$$
$\Rightarrow x^2+\frac{1}{x^2}=398$                               (Subtracting 2 from both sides)

Thus, the answer is 398.

Question 8:

If $x-\frac{1}{x}=3,$ find the values of $x^2+\frac{1}{x^2}$ and $x^4+\frac{1}{x^4}.$

Answer 8:

Let us consider the following equation:

$x-\frac{1}{x}=3$

Squaring both sides, we get:

$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^2={\left(3\right)}^2=9 \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=9 \\ \Rightarrow x^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=9 \\ \Rightarrow x^2-2+\frac{1}{x^2}=9 \end{gathered}$$
$\Rightarrow x^2+\frac{1}{x^2}=11$                               (Adding 2 to both sides)

Squaring both sides again, we get:

$$\begin{gathered} {\left(x^2+\frac{1}{x^2}\right)}^2={\left(11\right)}^2=121 \\ \Rightarrow {\left(x^2+\frac{1}{x^2}\right)}^2=121 \\ \Rightarrow {\left(x^2\right)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)+{\left(\frac{1}{x^2}\right)}^2=121 \\ \Rightarrow x^4+2+\frac{1}{x^4}=121 \end{gathered}$$
$\Rightarrow x^4+\frac{1}{x^4}=119$

Question 9:

If $x^2+\frac{1}{x^2}=18,$ find the values of $x+\frac{1}{x} \mathrm{and} x-\frac{1}{x}.$

Answer 9:

Let us consider the following expression:

$x+\frac{1}{x}$
Squaring the above expression, we get:

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2=x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2} \left[\right(a+b{)}^2=a^2+b^2+2ab] \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2=x^2+2+\frac{1}{x^2} \end{gathered}$$
$\Rightarrow {\left(x+\frac{1}{x}\right)}^2=20$                                                                     ($\because$ $x^2+\frac{1}{x^2}=18$)
$\Rightarrow x+\frac{1}{x}=\pm \sqrt{20}$                                                            (Taking square root of both sides)

Now, let us consider the following expression:

$x-\frac{1}{x}$
Squaring the above expression, we get:

$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^2=x^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2} \left[\right(a-b{)}^2=a^2+b^2-2ab] \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2} \end{gathered}$$
$\Rightarrow {\left(x-\frac{1}{x}\right)}^2=16$                                 ($\because$ $x^2+\frac{1}{x^2}=18$)
$\Rightarrow x-\frac{1}{x}=\pm 4$                                     (Taking square root of both sides)

Question 10:

If x + y = 4 and xy = 2, find the value of x² + y²

Answer 10:

We have:

$$\begin{gathered} {\left(x+y\right)}^2=x^2+2xy+y^2 \\ \Rightarrow x^2+y^2={\left(x+y\right)}^2-2xy \end{gathered}$$
$\Rightarrow x^2+y^2=4^2-2\times 2$                ($\because$ $x+y=4 \text{and} xy=2$)
$$\begin{gathered} \Rightarrow x^2+y^2=16-4 \\ \Rightarrow x^2+y^2=12 \end{gathered}$$

Question 11:

If x − y = 7 and xy = 9, find the value of x² + y²

Answer 11:

We have:

$$\begin{gathered} {\left(x-y\right)}^2=x^2-2xy+y^2 \\ \Rightarrow x^2+y^2={\left(x-y\right)}^2+2xy \end{gathered}$$
$\Rightarrow x^2+y^2=7^2+2\times 9$                     ($\because$ $x-y=7 \text{and} xy=9$ )
$$\begin{gathered} \Rightarrow x^2+y^2=7^2+2\times 9 \\ \Rightarrow x^2+y^2=49+18 \\ \Rightarrow x^2+y^2=67 \end{gathered}$$

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Question 12:

If 3x + 5y = 11 and xy = 2, find the value of 9x² + 25y²

Answer 12:

We have:

$$\begin{gathered} {\left(3x+5y\right)}^2={\left(3x\right)}^2+2\left(3x\right)\left(5y\right)+{\left(5y\right)}^2 \\ \Rightarrow {\left(3x+5y\right)}^2=9x^2+30xy+25y^2 \\ \Rightarrow 9x^2+25y^2={\left(3x+5y\right)}^2-30xy \end{gathered}$$
$\Rightarrow 9x^2+25y^2={11}^2-30\times 2$                          ($\because$ $3x+5y=11 \text{and} xy=2$)
$$\begin{gathered} \Rightarrow 9x^2+25y^2=121-60 \\ \Rightarrow 9x^2+25y^2=61 \end{gathered}$$

Question 13:

Find the values of the following expressions:
(i) 16x² + 24x + 9, when $x=\frac{7}{4}$
(ii) 64x² + 81y² + 144xy, when x = 11 and $y=\frac{4}{3}$
(iii) 81x² + 16y² − 72xy, when $x=\frac{2}{3}$ and $y=\frac{3}{4}$

Answer 13:

(i) Let us consider the following expression:

$16x^2+24x+9$

Now

$16x^2+24x+9={\left(4x+3\right)}^2$                                 (Using identity ${\left(a+b\right)}^2=a^2+2ab+b^2$)
$$\begin{gathered} \Rightarrow 16x^2+24x+9={\left(4\times \frac{7}{4}+3\right)}^2 (\mathrm{Substituting} x=\frac{7}{4}) \\ \Rightarrow 16x^2+24x+9={\left(7+3\right)}^2 \\ \Rightarrow 16x^2+24x+9={10}^2 \\ \Rightarrow 16x^2+24x+9=100 \end{gathered}$$


(ii) Let us consider the following expression:

$64x^2+81y^2+144xy$

Now

$64x^2+81y^2+144xy={\left(8x+9y\right)}^2$                                (Using identity ${\left(a+b\right)}^2=a^2+2ab+b^2$)
$$\begin{gathered} \Rightarrow 64x^2+81y^2+144xy={\left[8\left(11\right)+9\left(\frac{4}{3}\right)\right]}^2 (\mathrm{Substituting} x=11 \mathrm{and} y=\frac{4}{3}) \\ \Rightarrow 64x^2+81y^2+144xy={\left[88+12\right]}^2 \\ \Rightarrow 64x^2+81y^2+144xy={100}^2 \\ \Rightarrow 64x^2+81y^2+144xy=10000 \end{gathered}$$

(iii) Let us consider the following expression:

$81x^2+16y^2-72xy$

Now

$81x^2+16y^2-72xy={\left(9x-4y\right)}^2$                                  (Using identity ${\left(a+b\right)}^2=a^2-2ab+b^2$)
$$\begin{gathered} \Rightarrow 81x^2+16y^2-72xy={\left[9\left(\frac{2}{3}\right)-4\left(\frac{3}{4}\right)\right]}^2 (\mathrm{Substituting} x=\frac{2}{3}\mathrm{and} y=\frac{3}{4}) \\ \Rightarrow 81x^2+16y^2-72xy={\left[6-3\right]}^2 \\ \Rightarrow 81x^2+16y^2-72xy=3^2 \\ \Rightarrow 81x^2+16y^2-72xy=9 \end{gathered}$$

Question 14:

If $x+\frac{1}{x}=9,$ find the value of $x^4+\frac{1}{x^4}.$

Answer 14:

Let us consider the following equation:

$x+\frac{1}{x}=9$

Squaring both sides, we get:

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2={\left(9\right)}^2=81 \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2=81 \\ \Rightarrow x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=81 \\ \Rightarrow x^2+2+\frac{1}{x^2}=81 \end{gathered}$$
$\Rightarrow x^2+\frac{1}{x^2}=79$                               (Subtracting 2 from both sides)

Now, squaring both sides again, we get:

$$\begin{gathered} {\left(x^2+\frac{1}{x^2}\right)}^2={\left(79\right)}^2=6241 \\ \Rightarrow {\left(x^2+\frac{1}{x^2}\right)}^2=6241 \\ \Rightarrow {\left(x^2\right)}^2+2\left(x^2\right)\left(\frac{1}{x^2}\right)+{\left(\frac{1}{x^2}\right)}^2=6241 \\ \Rightarrow x^4+2+\frac{1}{x^4}=6241 \end{gathered}$$
$\Rightarrow x^4+\frac{1}{x^4}=6239$

Question 15:

If $x+\frac{1}{x}=12,$ find the value of $x-\frac{1}{x}.$

Answer 15:

Let us consider the following equation:

$x+\frac{1}{x}=12$

Squaring both sides, we get:

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2={\left(12\right)}^2=144 \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^2=144 \\ \Rightarrow x^2+2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=144 [ (a+b{)}^2=a^2+b^2+2ab] \\ \Rightarrow x^2+2+\frac{1}{x^2}=144 \end{gathered}$$
$\Rightarrow x^2+\frac{1}{x^2}=142$                               (Subtracting 2 from both sides)

Now

$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^2=x^2-2\times x\times \frac{1}{x}+{\left(\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2} \left[\right(a-b{)}^2=a^2+b^2-2ab] \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=x^2-2+\frac{1}{x^2} \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=142-2 (\because x^2+\frac{1}{x^2}=142) \\ \Rightarrow {\left(x-\frac{1}{x}\right)}^2=140 \\ \Rightarrow x-\frac{1}{x}=\pm \sqrt{140} \left(\text{Taking square root}\right) \end{gathered}$$

Question 16:

If 2x + 3y = 14 and 2x − 3y = 2, find the value of xy.
[Hint: Use (2x + 3y)² − (2x − 3y)² = 24xy]

Answer 16:

We will use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ to obtain the value of xy.
$\mathrm{Squaring} (2x+3y) \mathrm{and} (2x-3y) \mathrm{both} \mathrm{and} \mathrm{then} \mathrm{subtracting} \mathrm{them}, \mathrm{we} \mathrm{get}:$
$$\begin{gathered} {\left(2x+3y\right)}^2-{\left(2x-3y\right)}^2=\left\{\left(2x+3y\right)+\left(2x-3y\right)\right\}\left\{\left(2x+3y\right)-\left(2x-3y\right)\right\}=4x\times 6y=24xy \\ \Rightarrow {\left(2x+3y\right)}^2-{\left(2x-3y\right)}^2=24xy \end{gathered}$$
$$\begin{gathered} \Rightarrow 24xy={\left(2x+3y\right)}^2-{\left(2x-3y\right)}^2 \\ \Rightarrow 24xy={\left(14\right)}^2-{\left(2\right)}^2 \\ \Rightarrow 24xy=\left(14+2\right)\left(14-2\right) (\because \left(a+b\right)\left(a-b\right)=a^2-b^2) \\ \Rightarrow 24xy=16\times 12 \\ \Rightarrow xy=\frac{16\times 12}{24} \left(\text{Dividing both sides by }24\right) \\ \Rightarrow xy=8 \end{gathered}$$

Question 17:

If x² + y² = 29 and xy = 2, find the value of
(i) x + y
(ii) x − y
(iii) x⁴ + y⁴

Answer 17:

(i) We have:

$$\begin{gathered} {\left(x+y\right)}^2=x^2+2xy+y^2 \\ \Rightarrow \left(x+y\right)=\pm \sqrt{x^2+2xy+y^2} \\ \Rightarrow \left(x+y\right)=\pm \sqrt{29+2\times 2} ( \text{∵} x^2+y^2=29 \text{and} xy=2) \\ \Rightarrow \left(x+y\right)=\pm \sqrt{29+4} \\ \Rightarrow \left(x+y\right)=\pm \sqrt{33} \end{gathered}$$

(ii) We have:

$$\begin{gathered} {\left(x-y\right)}^2=x^2-2xy+y^2 \\ \Rightarrow \left(x-y\right)=\pm \sqrt{x^2-2xy+y^2} \\ \Rightarrow \left(x+y\right)=\pm \sqrt{29-2\times 2} (\text{∵} x^2+y^2=29 \text{and} xy=2) \\ \Rightarrow \left(x+y\right)=\pm \sqrt{29-4} \\ \Rightarrow \left(x+y\right)=\pm \sqrt{25} \\ \Rightarrow \left(x+y\right)=\pm 5 \end{gathered}$$

(iii) We have:

$$\begin{gathered} {\left(x^2+y^2\right)}^2=x^4+2x^2{y}^2+y^4 \\ \Rightarrow x^4+y^4={\left(x^2+y^2\right)}^2-2x^2{y}^2 \\ \Rightarrow x^4+y^4={\left(x^2+y^2\right)}^2-2{\left(xy\right)}^2 \\ \Rightarrow x^4+y^4={29}^2-2{\left(2\right)}^2 (\text{∵} x^2+y^2=29 \text{and} xy=2) \\ \Rightarrow x^4+y^4=841-8 \\ \Rightarrow x^4+y^4=833 \end{gathered}$$

Question 18:

What must be added to each of the following expressions to make it a whole square?
(i) 4x² − 12x + 7
(ii) 4x² − 20x + 20

Answer 18:

(i) Let us consider the following expression:

$4x^2-12x+7$
The above expression can be written as:

$4x^2-12x+7={\left(2x\right)}^2-2\times 2x\times 3+7$
It is evident that if 2x is considered as the first term and 3 is considered as the second term, 2 is required to be added to the above expression to make it a perfect square. Therefore, 7 must become 9.

Therefore, adding and subtracting 2 in the above expression, we get:

$\left(4x^2-12x+7\right)+2-2=\left\{{\left(2x\right)}^2-2\times 2x\times 3+7\right\}+2-2=\left\{{\left(2x\right)}^2-2\times 2x\times 3+9\right\}-2={\left(2x+3\right)}^2-2$
Thus, the answer is 2.

(ii) Let's consider the following expression:

$4x^2-20x+20$
The above expression can be written as:

$4x^2-20x+20={\left(2x\right)}^2-2\times 2x\times 5+20$
It is evident that if 2x is considered as the first term and 5 is considered as the second term, 5 is required to be added to the above expression to make it a perfect square. Therefore, number 20 must become 25.

Therefore, adding and subtracting 5 in the above expression, we get:

$\left(4x^2-20x+20+5\right)-5=\left\{{\left(2x\right)}^2-2\times 2x\times 5+20\right\}+5-5=\left\{{\left(2x\right)}^2-2\times 2x\times 5+25\right\}-5={\left(2x+5\right)}^2-5$
Thus, the answer is 5.

Question 19:

Simplify:
(i) (x − y)(x + y) (x² + y²)(x⁴ + y²)
(ii) (2x − 1)(2x + 1)(4x² + 1)(16x⁴ + 1)
(iii) (4m − 8n)² + (7m + 8n)²
(iv) (2.5p − 1.5q)² − (1.5p − 2.5q)²
(v) (m² − n²m)² + 2m³n²

Answer 19:

To simplify, we will proceed as follows:
(i) 
$$\begin{gathered} (i) \left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right) \\ =\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right) \left[\text{∵ }\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\left(x^4-y^4\right)\left(x^4+y^4\right) \left[\text{ ∵} \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =x^8-x^8 \left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \left(2x-1\right)\left(2x+1\right)\left(4x^2+1\right)\left(16x^4+1\right) \\ =\left({\left(2x\right)}^2-1^2\right)\left(4x^2+1\right)\left(16x^4+1\right) \left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\left(4x^2-1\right)\left(4x^2+1\right)\left(16x^4+1\right) \\ =\left\{{\left(4x^2\right)}^2-{\left(1^2\right)}^2\right\}\left(16x^4+1\right) \left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\left(16x^4-1\right) \left(16x^4+1\right) \\ ={\left(16x^4\right)}^2 - 1^2 \left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =256x^8-1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) {\left(7m-8n\right)}^2+{\left(7m+8n\right)}^2 \\ =2{\left(7m\right)}^2+2{\left(8n\right)}^2 \left[\text{∵ }{\left(a-b\right)}^2\text{+}{\left(a+b\right)}^2=2a^2+2b^2\right] \\ =98m^2+128n^2 \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iv}\right) {\left(2.5p-1.5q\right)}^2-{\left(1.5p-2.5q\right)}^2 \\ ={\left(2.5p\right)}^2+{\left(1.5q\right)}^2-2\left(2.5p\right)\left(1.5q\right)-\left[{\left(1.5p\right)}^2+{\left(2.5q\right)}^2-2\left(1.5p\right)\left(2.5q\right)\right] \\ ={\left(2.5p\right)}^2+{\left(1.5q\right)}^2\cancel{-2\left(2.5p\right)\left(1.5q\right)}-{\left(1.5p\right)}^2-{\left(2.5q\right)}^2+\cancel{2\left(1.5p\right)\left(2.5q\right)} \\ ={\left(2.5p\right)}^2-{\left(1.5p\right)}^2+{\left(1.5q\right)}^2-{\left(2.5q\right)}^2 \\ =\left[\left(2.5p+1.5p\right)\left(2.5p-1.5p\right)\right]+\left[\left(1.5q+2.5\right)\left(1.5q-2.5q\right)\right] \left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =4p\times p+4q\times \left(-q\right) \\ =4p^2-4q^2 \\ =4\left(p^2-4q^2 \right) \end{gathered}$$

$$\begin{gathered} \mathrm{v}) {\left(m^2-n^{2}m\right)}^2+2m^3{n}^2 \\ ={\left(m^2\right)}^2+{\left(n^{2}m\right)}^2 \left[\because {\left(a-b\right)}^2+2ab=a^2+b^2\right] \\ =m^4+n^4{m}^2 \end{gathered}$$

Question 20:

Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9a − 5b)² + 180ab = (9a + 5b)²
(iii) ${\left(\frac{4m}{3}-\frac{3n}{4}\right)}^2+2mn=\frac{16m^2}{9}+\frac{9n^2}{16}$
(iv) (4pq + 3q)² − (4pq − 3q)² = 48pq²
(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0

Answer 20:

$$\begin{gathered} \text{(i) LHS}={\left(3x+7\right)}^2-84x \\ ={\left(3x+7\right)}^2-4\times 3x\times 7 \\ ={\left(3x-7\right)}^2 \left[\because {\left(a+b\right)}^2-4ab={\left(a-b\right)}^2\right] \\ =\text{RHS} \\ \text{Because LHS is equal to RHS, the given equation is verified.} \end{gathered}$$


$$\begin{gathered} \text{(ii) LHS}={\left(9a-5b\right)}^2+180ab \\ ={\left(9a-5b\right)}^2+4\times 9a\times 5b \\ ={\left(9a+5b\right)}^2 \left[\because {\left(a-b\right)}^2+4ab={\left(a+b\right)}^2\right] \\ =\text{RHS} \\ \text{Because LHS is equal to RHS, the given equation is verified.} \end{gathered}$$


$$\begin{gathered} \text{(iii) LHS}={\left(\frac{4m}{3}-\frac{3n}{4}\right)}^2+2mn \\ ={\left(\frac{4m}{3}-\frac{3n}{4}\right)}^2+2\times \frac{4m}{3}\times \frac{3n}{4} \\ ={\left(\frac{4m}{3}\right)}^2+{\left(\frac{3n}{4}\right)}^2 \left[\because {\left(a-b\right)}^2+2ab=a^2+b^2\right] \\ =\frac{16m^2}{9}+\frac{9n^2}{16} \\ =\text{RHS} \\ \text{Because LHS is equal to RHS, the given equation is verified.} \end{gathered}$$


$$\begin{gathered} \text{(iv) LHS} \\ ={\left(4pq+3q\right)}^2-{\left(4pq-3q\right)}^2 \\ =4\left(4pq\right)\left(3q\right) \left[\because {\left(a+b\right)}^2-{\left(a+b\right)}^2=4ab\right] \\ =48p{q}^2 \\ =\text{RHS} \\ \text{Because LHS is equal to RHS, the given equation is verified.} \end{gathered}$$


$$\begin{gathered} \text{(v) LHS}=\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right) \\ =a^2-b^2+b^2-c^2+c^2-a^2 \left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\cancel{a^2}-\cancel{b^2}+\cancel{b^2}-\cancel{c^2}+\cancel{c^2}-\cancel{a^2 } \\ =0 \\ =\text{RHS} \\ \text{Because LHS is equal to RHS, the given equation is verified.} \end{gathered}$$