RD Sharma Solutions Class 8 Algebraic Expressions And Identities Exercise 6.7

RD Sharma Solutions Class 8 Chapter 6 Exercise 6.7

RD Sharma Class 8 Solutions Chapter 6 Ex 6.7 PDF Free Download

 Q1) Find the following products:

(i) (x + 4) (x + 7)

(ii) (x – 11) (x + 4)

(iii) (x + 7) (x – 5)

(iv) (x – 3) (x – 2)

(v) (y2 – 4) (y2 – 3)

(vi) \((x + \frac{4}{3})\; (x + \frac{3}{4})\)

(vii) (3x + 5) (3x + 11)

(viii) (2x2 – 3) (2x2 + 5)

(ix) (z2 + 2) (z2 – 3)

(x) (3x – 4y) (2x – 4y)

(xi) (3x2 – 4xy) (3x2 – 3xy)

(xii) \((x + \frac{1}{5})\; (x + 5)\)

(xiii) \((z + \frac{3}{4})\; (z + \frac{4}{3})\)

(xiv) (x2 + 4) (x2 + 9)

(xv) (y2 + 12) (y2 + 6)

(xvi) \((y^{2} + \frac{5}{7})\; (y^{2} – \frac{14}{5})\)

(xvii) \((p^{2} + 16)\; (p^{2} – \frac{1}{4})\)

Solution:

(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(x + 4) (x + 7)

= \(x^{2} + (4+7)x + 4\times 7\)

= \(x^{2} + 11x + 28\)

(ii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab.

(x – 11) (x + 4)

= \(x^{2} + (4-11)x – 11\times 4\)

= \(x^{2} – 7x – 44\)

(iii) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab.

(x + 7) (x – 5)

= \(x^{2} + (7-5)x – 7\times 5\)

= \(x^{2} + 2x – 35\)

(iv) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x + ab.

(x – 3) (x – 2)

= \(x^{2} – (3+2)x + 3\times 2\)

= \(x^{2} – 5x + 6\)

(v) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.

(y2 – 4) (y2 – 3)

= \((y^{2})^{2} – (4 + 3)(y^{2}) + 4\times 3\)

= \(y^{4} – 7y^{2} + 12\)

(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

\((x + \frac{4}{3})\; (x + \frac{3}{4})\)

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

= \(x^{2} + \frac{25}{12}x + 1\)

(vii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(3x + 5) (3x + 11)

= \((3x)^{2} + (5 + 11) (3x) + 5\times 11\)

= \(9x^{2} + 48x + 55\)

(viii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab.

(2x2 – 3) (2x2 + 5)

= \((2x^{2})^{2} + (5 – 3)(2x^{2}) – 3\times 5\)

= \(4x^{4} + 4x^{2} – 15\)

(ix) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab.

(z2 + 2) (z2 – 3)

= \((z^{2})^{2} + (2 – 3)(z^{2}) – 2\times 3\)

= \(z^{4} – z^{2} – 6\)

(x) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.

(3x – 4y) (2x – 4y)

= (4y – 3x) (4y – 2x)                           (Taking common -1 from both parentheses)

= \((4y)^{2} – (3x + 2x)\;(4y) + 3x\times 2x\)

= \(16y^{2} – (12xy + 8xy) + 6x^{2}\)

= \(16y^{2} – 20xy + 6x^{2}\)

(xi) Here, we will use the identity (x – a) (x – b) = x2 – (a + b)x – ab.

(3x2 – 4xy) (3x2 – 3xy)

= \((3x^{2})^{2} – (4xy + 3xy)\; (3x^{2}) + 4xy \times 3xy\)

= \(9x^{4} – (12x^{3}y + 9x^{3}y) + 12x^{2}y^{2}\)

= \(9x^{4} – 21x^{3}y + 12x^{2}y^{2}\)

(xii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

\((x + \frac{1}{5})\; (x + 5)\)

= \(x^{2} + (\frac{1}{5} + 5)x + \frac{1}{5}\times 5\)

= \(x^{2} + \frac{26}{5}x + 1\)

(xiii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

\((z + \frac{3}{4})\; (z + \frac{4}{3})\)

= \(z^{2} + (\frac{3}{4} + \frac{4}{3})x + \frac{3}{4}\times \frac{4}{3}\)

= \(z^{2} + \frac{25}{12}x + 1\)

(xiv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(x2 + 4) (x2 + 9)

= \((x^{2})^{2} + (4 + 9)(x^{2}) + 4 \times 9\)

= \(x^{4} + 13x^{2} + 36\)

(xv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab.

(y2 + 12) (y2 + 6)

= \((y^{2})^{2} + (12 + 6)(y^{2}) + 12 \times 6\)

= \(y^{4} + 18xy^{2} + 72\)

(xvi) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x –ab.

\((y^{2} + \frac{5}{7})\; (y^{2} – \frac{14}{5})\)

= \((y^{2})^{2} + (\frac{5}{7} – \frac{14}{5})(y^{2}) – \frac{5}{7}\times \frac{14}{5}\)

= \(y^{4} – \frac{73}{35}y^{2} – 2\)

(xvii) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x –ab.

\((p^{2} + 16)\; (p^{2} – \frac{1}{4})\)

= \((p^{2})^{2} + (16 – \frac{1}{4})(p^{2}) – 16\times \frac{1}{4}\)

= \(p^{4} + \frac{63}{4}p^{2} – 4\)

 

Q2. Evaluate the following:

(i) 102 x 106   

(ii) 109 x 107

(iii) 35 x 37

(iv) 53 x 55

(v) 103 x 96

(vi) 34 x 36

(vii) 994 x 1006

Solution:

(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

102 x 106

= (100 + 2) (100 + 6)

= 1002 + (2 + 6)100 + 2 x 6

=10000 + 800 +12 = 10812

(ii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

109 x 107

= (100 + 9) (100 + 7)

= 1002 + (9 + 7)100 + 9 x 7

=10000 + 1600 + 63 = 11663

(iii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

35 x 37

= (30 + 5) (30 + 7)

= 302 + (5 + 7)30 + 5 x 7

= 900 + 360 + 35 = 1295

(iv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

53 x 55

= (50 + 3) (50 + 5)

= 502 + (3 + 5)50 + 3 x 5

= 2500 + 400 + 15 = 2915

(v) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab

103 x 96

= (100 + 3) (100 – 4)

= 1002 + (3 – 4)100 – 3 x 4

= 10000 – 100 – 12 = 9888

(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab

34 x 36

= (30 + 4) (30 + 6)

= 302 + (4 + 6)30 + 4 x 6

= 900 + 300 + 24 = 1224

(vii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab

994 x 1006

= (1000 – 6) x (1000 + 6)

= 10002 + (6 – 6) x 1000 – 6 x 6

= 1000000 – 36 = 999964


Practise This Question

If the lowest temperature in Gulmarg in Kashmir was 10C in January and in summer the temperature rose by 43C and reached the maximum temperature, what was the maximum temperature that year?