Question 12 Marks
Simplify: $\sqrt{3+2\sqrt2}.$
AnswerWe are asked to simplify $\sqrt{3+2\sqrt2}.$ It can be written in the form $(\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}$ as$\sqrt{3+2\sqrt2}=\sqrt{2+1+2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2+2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2+1\big)^2}$
$=\sqrt2+1$
Hence the value of given expression is $\sqrt2+1.$
View full question & answer→Question 22 Marks
Write the rationalisation factor of $7-3\sqrt5.$
AnswerThe rationalizing factor of $\text{a}+\sqrt{\text{b}}$ is $\text{a}-\sqrt{\text{b}}.$ Hence the rationalizing factor of $7-3\sqrt5$ is $7+3\sqrt5.$
View full question & answer→Question 32 Marks
Retionalise the denominator of each of the following:$\frac{\sqrt2+\sqrt5}{\sqrt3}$
AnswerWe know that rationalization factor for $\frac{1}{\sqrt{\text{a}}}$ is $\sqrt{\text{a}}.$ we will multiply numerator and denominator of the given expression $\frac{\sqrt2+\sqrt5}{\sqrt3}$ by $\sqrt{3},$ to get$\frac{\sqrt2+\sqrt5}{\sqrt3}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{2}\times\sqrt3+\sqrt5\times\sqrt3}{\sqrt{3}\times\sqrt{3}}$
$=\frac{\sqrt{6}+\sqrt{15}}{3}$
Hence the given expression is simplified to $\frac{\sqrt{6}+\sqrt{15}}{3}.$
View full question & answer→Question 42 Marks
Find the value to three place of decimals of each of the following. It is given that $\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$$\frac{\sqrt{2}-1}{\sqrt5}$
AnswerGiven,$\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$
$\frac{\sqrt{2}-1}{\sqrt5}$
Rationalizing the denominator by multiplying both numerator and denominator with $\sqrt5$$=\frac{(\sqrt2\times\sqrt5)-\sqrt5}{\sqrt5\times\sqrt5}$
$=\frac{\sqrt{10}+\sqrt5}{5}$
$=\frac{3.162-2.236}{5}$
$=\frac{0.926}{5}=0.1852$
View full question & answer→Question 52 Marks
Find the value to three place of decimals of each of the following. It is given that $\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$$\frac{2+\sqrt{3}}{3}$
AnswerGiven,$\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$
$\frac{2+\sqrt{3}}{3}$
$=\frac{2+1.732}{3}$
$=\frac{3.732}{3}$
$=1.244$
View full question & answer→Question 62 Marks
Simplify: $\sqrt{3-2\sqrt2}.$
AnswerWe are asked to simplify $\sqrt{3-2\sqrt2}.$ It can be written in the form $(\text{a}-\text{b})^2=\text{a}^2+\text{b}^2-2\text{ab}$ as$\sqrt{3-2\sqrt2}=\sqrt{2+1-2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2-2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2-1\big)^2}$
$=\sqrt2-1$
Hence the value of given expression is $\sqrt2-1.$
View full question & answer→Question 72 Marks
Simplify the following expression:$(4+\sqrt{7})(3+\sqrt{2})$
Answer$(4+\sqrt{7})(3+\sqrt{2})$$=12+4\sqrt{2}+3\sqrt{7}+\sqrt{7\times2}$
$=12+4\sqrt{2}+3\sqrt{7}+\sqrt{14}$
View full question & answer→Question 82 Marks
Find the value to three place of decimals of each of the following. It is given that $\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$$\frac{2}{\sqrt3}$
AnswerGiven,$\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$
$\frac{2}{\sqrt3}$
Rationalizing the denominator by multiplying both numerator and denominator with$=\frac{2\sqrt3}{\sqrt3\times\sqrt3}$
$=\frac{2\sqrt3}{\sqrt{3\times3}}$
$=\frac{2\sqrt3}{3}$
$=\frac{2\times1.732}{3}$
$=\frac{3.464}{3}=1.154666666$
View full question & answer→Question 92 Marks
Simplify the following expression:$(\sqrt{8}-\sqrt{2})(\sqrt{8}+\sqrt{2})$
Answer$(\sqrt{8}-\sqrt{2})(\sqrt{8}+\sqrt{2})$As we know, $(a + b)(a - b) = (a^2 - b^2)$
$\sqrt{8\times8}-\sqrt{2\times2}=8-2$
$=6$
View full question & answer→Question 102 Marks
Simplify the following:$\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}$
AnswerWe know that $\frac{\sqrt[\text{n}]{\text{a}}}{{\sqrt[\text{n}]{\text{b}}}}=\sqrt[\text{n}]{\frac{\text{a}}{\text{b}}}.$ We will use this property to simplify the expression $\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}.$$\therefore\ \frac{\sqrt[4]{1250}}{\sqrt[4]{2}}=\sqrt[4]{625}$
$=\sqrt[4]{5^4}$
$=(5^4)^\frac{1}{4}$
$=(5)^1$
$=5$
Hence the value of the given expression is 5.
View full question & answer→Question 112 Marks
Write the rationalisation factor of $\sqrt5-2.$
AnswerGiven that $\sqrt5-2,$ we know that rationalization factor of $\sqrt{\text{a}}-\text{b}$ is $\sqrt{\text{a}}+\text{b}$
So the rationalization factor of $\sqrt5-2$ is $\sqrt5+2.$
View full question & answer→Question 122 Marks
Express each one of the following with rational denominator:$\frac{1}{3+\sqrt2}$
Answer$\frac{1}{3+\sqrt2}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $3-\sqrt2$
$=\frac{3-\sqrt2}{(3+\sqrt2)(3-\sqrt2)}$
As we know,
$(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)=\frac{3-\sqrt2}{9-2}=\frac{3-\sqrt2}{7}$
View full question & answer→Question 132 Marks
Simplify the following:$\sqrt[3]{4}\times\sqrt[3]{16}$
AnswerWe know that $\sqrt[\text{n}]{\text{a}}\times\sqrt[\text{n}]{\text{b}}=\sqrt[\text{n}]{\text{ab}}.$ We will use this property to simplify the expression $\sqrt[3]{4}\times\sqrt[3]{16.}$$\therefore\ \sqrt[3]{4}\times\sqrt[3]{16.}=\sqrt[3]{64}$
$=\sqrt[3]{4^3}$
$=(4^3)^\frac{1}{3}$
$=(4)^1$
$=4$
Hence the value of the given expression is 4.
View full question & answer→Question 142 Marks
Simplify the following expression:$(3+\sqrt{3})(3-\sqrt{3})$
Answer$(3+\sqrt{3})(3-\sqrt{3})$As we know, $(a + b)(a - b) = (a^2 - b^2)$
$=9-\sqrt{3\times3}$
$=6$
View full question & answer→Question 152 Marks
Simplify the following expression:$(5+\sqrt{7})(5-\sqrt{7})$
Answer$(5+\sqrt{7})(5-\sqrt{7})$
As we know, $(a+b)(a-b)=\left(a^2-b^2\right)$
So, $5^2 - 7$
$25 - 7 = 18$
View full question & answer→Question 162 Marks
Simplify the following expression:$(3+\sqrt{3})(5-\sqrt{2})$
Answer$(3+\sqrt{3})(5-\sqrt{2})$$=\sqrt{15}-3\sqrt{2}+5\sqrt{3}-\sqrt{3\times2}$
$=\sqrt{15}-3\sqrt{2}+5\sqrt{3}-\sqrt{6}$
View full question & answer→Question 172 Marks
Retionalise the denominator of each of the following:$\frac{3}{2\sqrt5}$
AnswerWe know that rationalization factor for $\frac{1}{\sqrt{\text{a}}}$ is $\sqrt{\text{a}}.$ we will multiply numerator of the given expression $\frac{3}{2\sqrt5}$ by $\sqrt5,$ to get$\frac{3}{2\sqrt5}\times\frac{\sqrt5}{\sqrt5}=\frac{3\sqrt5}{2\sqrt5\times\sqrt5}$
$=\frac{3\sqrt5}{2\times5}$
$=\frac{3\sqrt5}{10}$
Hence the given expression is simplified to $\frac{3\sqrt5}{10}.$
View full question & answer→Question 182 Marks
Express each one of the following with rational denominator:$\frac{1}{2\sqrt{5}-\sqrt3}$
Answer$\frac{1}{2\sqrt{5}-\sqrt3}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $2\sqrt{5}+\sqrt3$
$=\frac{2\sqrt5+\sqrt3}{(2\sqrt{5}-\sqrt3)(2\sqrt{5}+\sqrt3)}$
As we know, $(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{2\sqrt5+\sqrt3}{20-3}=\frac{2\sqrt5+\sqrt3}{17}$
View full question & answer→Question 192 Marks
Simplify the following expression:$(11+\sqrt{11})(11-\sqrt{11})$
Answer$(11+\sqrt{11})(11-\sqrt{11})$ As we know, $(a + b)(a - b) = (a^2 - b^2)$
So, $11^2 - 11$
$121 - 11 = 110$
View full question & answer→Question 202 Marks
Retionalise the denominator of each of the following:$\frac{\sqrt2}{\sqrt5}$
AnswerWe know that rationalization factor for $\frac{1}{\sqrt{\text{a}}}$ is $\sqrt{\text{a}}.$ we will multiply numerator and denominator of the given expression $\frac{\sqrt2}{\sqrt5}$ by $\sqrt{5},$ to get$\frac{\sqrt2}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{2}\times\sqrt5}{\sqrt{5}\times\sqrt{5}}$
$=\frac{\sqrt{10}}{5}$
Hence the given expression is simplified to $\frac{\sqrt10}{5}.$
View full question & answer→Question 212 Marks
Simplify the following expression:$(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$
Answer$(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$As we know, $(a + b)(a - b) = (a^2 - b^2)$
$=\sqrt{5\times5}-\sqrt{2\times2}$
$=5-2$
$=3$
View full question & answer→Question 222 Marks
Find the value to three place of decimals of each of the following. It is given that $\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$$\frac{3}{\sqrt{10}}$
AnswerGiven,$\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$
$\frac{3}{\sqrt{10}}$
Rationalizing the denominator by multiplying both numerator and denominator with $\sqrt{10}$$=\frac{3\sqrt{10}}{\sqrt{10}\times\sqrt{10}}$
$=\frac{3\sqrt{10}}{\sqrt{10\times10}}$
$=\frac{3\sqrt10}{10}$
$=\frac{9.486}{10}=0.9486$
View full question & answer→Question 232 Marks
Simplify the following expression:$(2\sqrt{5}+3\sqrt{2})^2$
Answer$(2 \sqrt{5}+3 \sqrt{2})^2$ As we know, $(a+b)^2=\left(a^2+2 \times b+b^2\right)$
$=4\sqrt{5\times5}+2\times2\sqrt{5}\times3\sqrt{2}+9\sqrt{2\times2}$
$=20+12\sqrt{10}+18$
$=38+12\sqrt{10}$
View full question & answer→Question 242 Marks
Retionalise the denominator of each of the following:$\frac{\sqrt3+1}{2}$
AnswerWe know that rationalization factor for $\frac{1}{\sqrt{\text{a}}}$ is $\sqrt{\text{a}}.$ we will multiply numerator and denominator of the given expression $\frac{\sqrt3+1}{2}$ by $\sqrt{2},$ to get$\frac{\sqrt3+1}{2}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}\times\sqrt3+\sqrt2}{\sqrt{2}\times\sqrt{2}}$
$=\frac{\sqrt{6}+\sqrt2}{2}$
Hence the given expression is simplified to $\frac{\sqrt{6}+\sqrt2}{2}.$
View full question & answer→Question 252 Marks
If $\text{x}=3+2\sqrt2,$ than find the value of $\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}.$
AnswerWe are asked to simplify $\text{x}=3+2\sqrt2.$ It can be written in the form $(\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}$ as$\sqrt{\text{x}}=\sqrt{3+2\sqrt2}$
$=\sqrt{2+1+2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2+2\times1\times\sqrt2}$
$=\sqrt{\big(\sqrt2+1\big)^2}$
$=\sqrt2+1$
Therefore,$\frac{1}{\sqrt{\text{x}}}=\frac{1}{\sqrt2+1}$
We know that rationalization factor for $\sqrt2+1$ is $\sqrt2-1.$ We will multiply numerator and denominator of the given expression $\frac{1}{\sqrt2+1}$ by $\sqrt2-1,$ to get$\frac{1}{\sqrt2+1}\times\frac{\sqrt2-1}{\sqrt2-1}=\frac{\sqrt2-1}{\big(\sqrt2\big)^2-(1)^2}$
$=\frac{\sqrt2-1}{2-1}$
$=\sqrt2-1$
Hence,$\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}=\sqrt2+1-\big(\sqrt2-1\big)$
$\sqrt2+1-\sqrt2+1$
$=2$
Therefore, value of given expression is 2.
View full question & answer→Question 262 Marks
Simplify the following expression:$(\sqrt{5}-2)(\sqrt{3}-\sqrt{5})$
Answer$(\sqrt{5}-2)(\sqrt{3}-\sqrt{5})$$=\sqrt{15}-\sqrt{25}-2\sqrt{3}+2\sqrt{5}$
$=\sqrt{15}-\sqrt{5\times5}-2\sqrt{3}+2\sqrt{5}$
$=\sqrt{15}-5-2\sqrt{3}+2\sqrt{5}$
View full question & answer→Question 272 Marks
Express each one of the following with rational denominator:$\frac{1}{\sqrt6-\sqrt5}$
AnswerWe have $\frac{1}{\sqrt6-\sqrt5}.$ Rationalisation factor for $\frac{1}{\sqrt6-\sqrt5}$ is $\sqrt6+\sqrt5$$\Rightarrow\frac{1}{\sqrt6-\sqrt5}=\frac{1}{\sqrt6-\sqrt5}\times\frac{\sqrt6+\sqrt5}{\sqrt6+\sqrt5}$
$=\frac{\sqrt6+\sqrt5}{(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)}$
$=\frac{\sqrt6+\sqrt5}{(\sqrt6)^2-(\sqrt5)^2}$ $[\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$
$=\frac{\sqrt6+\sqrt5}{6-5}$
$=\frac{\sqrt6+\sqrt5}{1}=\sqrt6+\sqrt5$
$\therefore\frac{1}{\sqrt6-\sqrt5}=\sqrt6+\sqrt5$
View full question & answer→Question 282 Marks
Retionalise the denominator of each of the following:$\frac{3\sqrt2}{\sqrt5}$
AnswerWe know that rationalization factor for $\frac{1}{\sqrt{\text{a}}}$ is $\sqrt{\text{a}}.$ we will multiply numerator and denominator of the given expression $\frac{3\sqrt2}{\sqrt5}$ by $\sqrt{5},$ to get$\frac{3\sqrt2}{\sqrt5}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{3\sqrt{2}\times\sqrt5}{\sqrt{5}\times\sqrt{5}}$
$=\frac{3\sqrt{10}}{5}$
Hence the given expression is simplified to $\frac{3\sqrt{10}}{5}.$
View full question & answer→Question 292 Marks
Express each one of the following with rational denominator:$\frac{\sqrt3+1}{2\sqrt{2}-\sqrt3}$
Answer$\frac{\sqrt3+1}{2\sqrt{2}-\sqrt3}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $2\sqrt{2}+\sqrt3$
$=\frac{(\sqrt3+1)(2\sqrt2+\sqrt3)}{(2\sqrt{2}+\sqrt3)(2\sqrt{2}-\sqrt3)}$
As we know, $(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{(2\sqrt6+3+2\sqrt2+\sqrt3)}{8-3}$
$=\frac{(2\sqrt6+3+2\sqrt2+\sqrt3)}{5}$
View full question & answer→Question 302 Marks
Write the reciprocal $5+\sqrt2.$
AnswerGiven that $5+\sqrt2,$ it's reciprocal is given as$\frac{1}{5+\sqrt2}$
It can be simplified by rationalizing the denominator. The rationalizing factor of $5+\sqrt2$ is $5-\sqrt2,$ we will multiply numerator and denominator of the given expression $\frac{1}{5+\sqrt2}$ by $5-\sqrt2,$ to get$\frac{1}{5+\sqrt2}\times\frac{5-\sqrt2}{5-\sqrt2}=\frac{5-\sqrt2}{(5)^2\big(\sqrt2\big)^2}$
$=\frac{5-\sqrt2}{25-2}$
$=\frac{5-\sqrt2}{23}$
Hence reciprocal of the given expression is $\frac{5-\sqrt2}{23}.$
View full question & answer→Question 312 Marks
Simplify the following expression:$(\sqrt{5}-\sqrt{3})^2$
Answer$(\sqrt{5}-\sqrt{3})^2$ As we know, $(a-b)^2=\left(a^2-2 \times a \times b+b^2\right)$
$=(\sqrt{5})^2-2\times\sqrt{5\times3}+(\sqrt{3})^2$
$=5-2\sqrt{15}+3$
$=8-2\sqrt{15}$
View full question & answer→Question 322 Marks
Simplify the following expression:$(\sqrt{3}+\sqrt{7})^2$
Answer$(\sqrt{3}+\sqrt{7})^2$As we know, $(a + b)^2= (a^2 + 2 \times a \times b + b^2)$
$=\sqrt{3^2}+2\times\sqrt{3}\times\sqrt{7}+\sqrt{7^2}$
$=3+2\times\sqrt{3\times7}+7$
$=10+2\times\sqrt{21}$
View full question & answer→Question 332 Marks
Find the value to three place of decimals of each of the following. It is given that $\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$$\frac{\sqrt5+1}{\sqrt{2}}$
AnswerGiven,$\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$
$\frac{\sqrt5+1}{\sqrt{2}}$
Rationalizing the denominator by multiplying both numerator and denominator with $\sqrt2$$=\frac{(\sqrt{5}\times\sqrt2)+\sqrt2}{\sqrt{2}\times\sqrt{2}}$
$=\frac{\sqrt{10}+\sqrt2}{2}$
$=\frac{4.576}{2}=2.288$
View full question & answer→Question 342 Marks
Retionalise the denominator of each of the following:$\frac{1}{\sqrt12}$
AnswerWe know that rationalization factor for $\frac{1}{\sqrt{\text{a}}}$ is $\sqrt{\text{a}}.$ we will multiply numerator and denominator of the given expression $\frac{1}{\sqrt{12}}$ by $\sqrt{12},$ to get$\frac{1}{\sqrt{12}}\times\frac{\sqrt{12}}{\sqrt{12}}=\frac{\sqrt{12}}{\sqrt{12}\times\sqrt{12}}$
$=\frac{\sqrt{12}}{12}$
$=\frac{\sqrt{4}\times\sqrt3}{12}$
$=\frac{2\times\sqrt3}{12}$
$=\frac{\sqrt3}{6}$
Hence the given expression is simplified to $\frac{\sqrt3}{6}.$
View full question & answer→Question 352 Marks
Write the value of $\big(2+\sqrt3)(2-\sqrt3\big).$
AnswerGiven that:$\big(2+\sqrt3)(2-\sqrt3\big)$
It can be simplified as$\big(2+\sqrt3)(2-\sqrt3\big)=2\times2-2\times\sqrt3+2\times\sqrt3-\big(\sqrt3\big)^2$
$=4-2\sqrt3+2\sqrt3-3$
$=4-3$
$=1$
Hence the value of the given expression is 1.
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