Advertisement:


Exercise 4.1

On the following page we have given the solution of Exercise 4.1 of Mathematics 9 (Science) published by Caravan Book House, Lahore.

We have created this page and it will be updated to add new solutions occasionally. Please stay in touch with this page.

Question 1:

Identify whether the following algebraic expressions are polynomials (Yes or No) (i) $3x^2+\frac{1}{x}-5$

(ii) $3x^3-4x^2-x\sqrt{x}+3$

(iii) $x^2-3x+\sqrt{2}$

(iv) $\frac{3x}{2x-1}+8$

Solution:

(i) $3x^2+\frac{1}{x}-5$
$No (Reason:\frac{1}{x})$
(ii) $3x^3-4x^2-x\sqrt{x}+3$
$No (Reasons \sqrt{x})$
(iii) $x^2-3x+\sqrt{2}$
$Yes$
(iv) $\frac{3x}{2x-1}+8$
$No (Reason:\frac{1}{2x-1})$

Question 2:

State whether each of the following expression is a rational expresion or not.
(i) $\frac{3 \sqrt{x}}{3 \sqrt{x}+5}$

(ii) $\frac{x^3-2x^2+\sqrt{3}}{2+3x-x^2}$

(iii) $\frac{x^2+6x+9}{x^2-9}$

(iv) $\frac{2\sqrt{x}+3}{2\sqrt{x}-3}$

Solution:

(i) $3x^2+\frac{1}{x}-5$
$No (Reason:\frac{1}{x})$
(ii) $3x^3-4x^2-x\sqrt{x}+3$
$No Reasons \sqrt{x}$
(iii) $x^2-3x+\sqrt{2}$
$Yes$
(iv) $\frac{3x}{2x-1}+8$
$No (Reason:\frac{1}{2x-1})$

Question 3:

Reduce the following rational expressions to the lowest forms.

(i) $\frac{120 x^2y^3z^5}{30x^3yz^2}$

(ii) $\frac{8a(x+1)}{2(x^2-1)}$
(iii) $\frac{(x+y)^2-4xy}{(x-y)^2}$
(iv) $\frac{(x^3-y^3)(x^2-2xy+y^2)}{(x-y)(x^2+xy+y^2)}$

(v) $\frac{(x+2)(x^2-1)}{(x+1)(x^2-4)}$
(vi) $\frac{x^2-4x+4}{2x^2-8}$
(vii) $\frac{64x^5-64x}{(8x^2+8)(2x+2)}$
(viii) $\frac{9x^2-(x^2-4)^2}{4+3x-x^2}$

Solution:
(i) $\frac{120 x^2y^3z^5}{30x^3yz^2}$
$\begin{align}\frac{30\times 4 y^(3-1)z^(5-2)}{30x^(3-2)}\\&=\frac{4 y^2 z^3}{x}\end{align}$
Solution:
(ii) $\frac{8 a(x+1)}{2(x^2-1)}$
$\begin{align}\frac{2\times 4a(x+1)}{2(x+1)(x-1)}\\&= \frac{4a}{x-1}\end{align}$
Solution:
(iii) $\frac{(x+y)^2-4xy}{(x-y)^2}$
$\begin{align}\frac{x^2+y^2+2xy-4xy}{x^2-2xy+y^2}\\&= 1\end{align}$
Solution:
(iv) $\frac{(x^3-y^3)(x^2-2xy+y^2)}{(x-y)(x^2+xy+y^2)}$
$\begin{align}\frac{(x^3-y^3)(x^2-2xy+y^2)}{(x^3-y^3)}\\&= x^2-2x+y^2\\&=(x-y)^2\end{align}$
Solution:
(v) $\frac{(x+2)(x^2-1)}{(x+1)(x^2-4)}$
$\begin{align}\frac{(x+2)(x^2-1)}{(x+1)(x^2-2^2)}\\&= \frac{(x+2)(x+1)(x-1)}{(x+1)(x+2)(x-2)}\\&=\frac{x-1}{x-2}\end{align}$
Solution:
(vi) $\frac{x^2-4x+4}{2x^2-8}$
$\begin{align}\frac{x^2-2x-2x+4}{2(x^2-4)}\\&= \frac{x(x-2)-2(x-2)}{2(x^2-4)}\\&=\frac{(x-2)(x-2)}{(x-2)(x+2)}\\&= \frac{x-2}{2(x+2)}\end{align}$
Solution:
(vii) $\frac{64x^5-64x}{(8x^2+8)(2x+2)}$
$\begin{align}\frac{64x(x^4-1)}{8(x^2+1)2(x+1)}\\&= \frac{64x((x^2)^2-1)}{16(x^2+1)(x+1)}\\&=\frac{64x(x^2+1)(x^2-1)}{16(x^2+1)(x+1)}\\&= \frac{64x(x^2+1)(x-1)(x+1)}{16(x^2+1)(x+1)}\\&= 4x(x-1)\end{align}$
Solution:
(viii) $\frac{9x^2-(x^2-4)^2}{4+3x-x^2}$
$\begin{align}\frac{(3x)^2-(x^2-4)^2}{4+3x-x^2}\\&= \frac{(3x-(x^2-4))(3x+(x^2-4))}{4+3x-x^2}\\&= \frac{(3x-x^2+4)(3x+x^2-4)}{4+3x-x^2}\\&= \frac{(x^2+3x-4)(4+3x-x^2)}{4+3x-x^2}\\&= x^2+3x-4)\end{align}$


Advertisement: