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

AcademicMathematicsNCERTClass 8

To do:

We have to write the given squares of bionomials as trinomials.

Solution:

We know that,

$(a + b)^2 = a^2 + 2ab + b^2$

$(a - b)^2 = a^2 - 2ab + b^2$

Therefore,

(i) $(x + 2)^2 = (x)^2 + 2(x)(2) + (x)^2$

$= x^2 + 4x + 4$

(ii) $(8a + 3b)^2 = (8a)^2 + 2(8a)(3b) + (3b)^2$

$= 64a^2 + 48ab + 9b^2$

(iii) $(2m + 1)^2 = (2m)^2 + 2(2m)(1) + (1)^2$

$= 4m^2 + 4m + 1$

(iv) $(9a + \frac{1}{6})^2 = (9a)^2 + 2(9a)(\frac{1}{6}) + (\frac{1}{6})^2$

$= 81a^2 + 3a + \frac{1}{36}$

(v) $(x + \frac{x^2}{2})^2 = (x)^2 + 2(x)(\frac{x^2}{2}) + (\frac{x^2}{2})^2$

$= x^2 + x^{1+2} + \frac{(x^2)^2}{4}$

$=x^2+x^3+\frac{x^4}{4}$

(vi) $(\frac{x}{4} - \frac{y}{3})^2 = (\frac{x}{4})^2 - 2(\frac{x}{4})(\frac{y}{3}) + (\frac{y}{3})^2$

$= \frac{x^2}{16} - \frac{xy}{6} + \frac{y^2}{9}$

(vii) $(3x - \frac{1}{3x})^2 = (3x)^2 - 2(3x)(\frac{1}{3x}) + (\frac{1}{3x})^2$

$= 9x^2 - 2 + \frac{1}{9x^2}$

(viii) $(\frac{x}{y} - \frac{y}{x})^2 = (\frac{x}{y})^2 - 2(\frac{x}{y})(\frac{y}{x}) + (\frac{y}{x})^2$

$= \frac{x^2}{y^2} - 2 + \frac{y^2}{x^2}$

(ix) $(\frac{3a}{2} - \frac{5b}{4})^2 = (\frac{3a}{2})^2 - 2(\frac{3a}{2})(\frac{5b}{4}) + (\frac{5b}{4})^2$

$= \frac{9a^2}{4} - \frac{15ab}{4} + \frac{25b^2}{16}$

(x) $(a^2b-bc^2)^2 = (a^2b)^2 - 2(a^2b)(bc^2) + (bc^2)^2$

$= a^4b^2 - 2a^2b^2c^2 + b^2c^4$

(xi) $(\frac{2 a}{3 b}+\frac{2 b}{3 a})^2 = (\frac{2a}{3b})^2 + 2(\frac{2a}{3b})(\frac{2b}{3a}) + (\frac{2b}{3a})^2$

$= \frac{4a^2}{9b^2} + \frac{8}{9} + \frac{4b^2}{9a^2}$ 

(xii) $(x^2-ay)^2 = (x^2)^2 - 2(x^2)(ay) + (ay)^2$

$= x^4 - 2ax^2y + a^2y^2$

raja
Updated on 10-Oct-2022 13:19:30

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