Exercise 3.7 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.7$
Question $1 .$
Find the square root of the following rational expressions.
(i) $\frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}}$
(ii) $\frac{7 x^{2}+2 \sqrt{14} x+2}{x^{2}-\frac{1}{2} x+\frac{1}{16}}$
(iii) $\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}}$
Solution:
(i) $\sqrt{\frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}}}=\frac{20 x^{2} y^{6} z^{8}}{10 x^{4} y^{2} z^{2}}$
$=\frac{2\left|y^{4} z^{6}\right|}{\left|x^{2}\right|}$
(ii) $\sqrt{\frac{7 x^{2}+2 \sqrt{14} x+2}{x^{2}-\frac{1}{2} x+\frac{1}{16}}}=\sqrt{\frac{(\sqrt{7} x+\sqrt{2})(\sqrt{7} x+\sqrt{2})}{\left(x-\frac{1}{4}\right)\left(x-\frac{1}{4}\right)}}$
$=4\left|\frac{(\sqrt{7} x+\sqrt{2})}{4 x-1}\right| \quad \begin{aligned}&=\frac{\sqrt{2 \times 7}}{\sqrt{7} \times \sqrt{7}} \quad \frac{\sqrt{2 \times 7}}{\sqrt{7} \times \sqrt{7}} \\&=(\sqrt{7} x+\sqrt{2})(\sqrt{7} x+\sqrt{2})\end{aligned}$
(iii) $\sqrt{\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}}}$

(iii) $\sqrt{\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}}}$
$=\frac{11(a+b)^{4}(x+y)^{4}(b-c)^{4}}{9(b-c)^{2}(a-b)^{6}(b-c)^{2}}$
$=\frac{11}{9}\left|\frac{(a+b)^{4}(x+y)^{4}}{(a-b)^{6}}\right|$

Question $2 .$
Find the square root of the following
(i) $4 x^{2}+20 x+25$
(ii) $9 x^{2}-24 x y+30 x z-40 y z+25 z^{2}+16 y^{2}$
(iii) $1+\frac{1}{x^{6}}+\frac{2}{x^{3}}$
(iv) $\left(4 \mathrm{x}^{2}-9 \mathrm{x}+2\right)\left(7 \mathrm{x}^{2}-13 \mathrm{x}-2\right)\left(28 \mathrm{x}^{2}-3 \mathrm{x}-1\right)$
(v) $\left(2 \mathrm{x}^{2}+\frac{17}{6} x+1\right)\left(\frac{3}{2} x^{2}+4 \mathrm{x}+2\right)\left(\frac{4}{3} x^{2}+\frac{11}{3} x+2\right)$
Solution:

(i) $\sqrt{4 x^{2}+20 x+25}=\sqrt{(2 x+5)^{2}}=|2 x+5|$
(ii) $\sqrt{9 x^{2}-24 x y+30 x z-40 y z+25 z^{2}+16 y^{2}}$
(iii) $\sqrt{1+\frac{1}{x^{6}}+\frac{2}{x^{3}}}=\sqrt{1^{2}+2 \cdot 1 \cdot \frac{1}{x^{3}}+\left(\frac{1}{x^{3}}\right)^{2}}$
$=\sqrt{\left(1+\frac{1}{x^{3}}\right)^{2}}=\left|1+\frac{1}{x^{3}}\right|$
(iv)
$\begin{aligned}
& \sqrt{\left(4 x^{2}-9 x+2\right)\left(7 x^{2}-13 x-2\right)\left(28 x^{2}-3 x-1\right)} \\
=& \sqrt{(x-2)(4 x-1)(x-2)(7 x+1)(4 x-1)(7 x+1)} \\
=& \sqrt{(x-2)^{2}(4 x-1)^{2}(7 x+1)^{2}} \\
=&|(x-2)(4 x-1)(7 x+1)|
\end{aligned}$
(v)
$\begin{aligned}
&\left.\sqrt{\left(2 x^{2}+\frac{17}{6} x+1\right)\left(\frac{3}{2} x^{2}+4 x+2\right)\left(\frac{4}{3} x^{2}+\frac{11}{3} x+2\right)}\right) \\
&=\sqrt{\frac{\left(12 x^{2}+17 x+6\right)}{6}\left(\frac{3 x^{2}+8 x+4}{2}\right)\left(\frac{4 x^{2}+11 x+6}{3}\right)}
\end{aligned}$

$=\frac{1}{6} \sqrt{(4 x+3)(3 x+2)(x+2)(3 x+2)(4 x+3)(x+2)}$
$=\frac{1}{6}|(4 x+3)(3 x+2)(x+2)|$