Question 13 Marks
If $\frac{\sqrt3-1}{\sqrt3+1}=\text{x}+\text{y}\sqrt3,$ find the value of x and y.
AnswerIt is given that:$\frac{\sqrt3-1}{\sqrt3+1}=\text{x}+\text{y}\sqrt3.$ we need to find x and y.
We know that rationalization factor for $\sqrt3+1$ is $\sqrt3-1.$ We will multiply numerator and denominator of the given expression $\frac{\sqrt3-1}{\sqrt3+1}$ by $\sqrt3-1,$ to get$\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}=\frac{\big(\sqrt3\big)^2+(1)^2-2\times\sqrt3\times1}{\big(\sqrt3\big)^2-(1)^2}$
$=\frac{3+1-2\sqrt3}{3-1}$
$=\frac{4-2\sqrt3}{2}$
$=2-\sqrt3$
On equating rational and irrational terms, we get$\text{x}+\text{y}\sqrt3=2-\sqrt3$
Hence, we get x = 2, y = -1.
View full question & answer→Question 23 Marks
Rationales the denominator and simplify:$\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
Answer$\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt3-\sqrt2$
$=\frac{\big(\sqrt3-\sqrt2\big)\big(\sqrt3-\sqrt2\big)}{\big(\sqrt3+\sqrt2\big)\big(\sqrt3-\sqrt2\big)}$
As we know, $(\text{a}-\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}$
As we know, $(\text{a}-\text{b})^2=(\text{a}^2-2\times\text{a}\times\text{b}+\text{b}^2)$
$=\frac{3-2\sqrt3\sqrt2+2}{1}=5-2\sqrt6$
View full question & answer→Question 33 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.$$\frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$
AnswerGiven,$\sqrt2=1.414,\ \sqrt3=1.732,\ \sqrt5=2.236,\ \sqrt10=3.162.$
$\frac{\sqrt{10}+\sqrt{15}}{\sqrt{2}}$
Rationalizing the denominator by multiplying both numerator and denominator with $\sqrt{2}$$=\frac{(\sqrt{10}\times\sqrt2)+(\sqrt{15}\times\sqrt2)}{\sqrt{2}\times\sqrt{2}}$
$=\frac{\sqrt{20}+\sqrt{30}}{2}$
$=\frac{(\sqrt{10}\times\sqrt2)+(\sqrt{10}\times\sqrt3)}{2}$
$=\frac{(3.162\times1.414)+(3.162\times1.732)}{2}$
$=\frac{(4.471068)+(5.476584)}{2}$
$=\frac{9.947652}{2}$
$=4.973826$
View full question & answer→Question 43 Marks
Express each one of the following with rational denominator:$\frac{\text{b}^2}{\sqrt{\text{a}^2+\text{b}^2}+\text{a}}$
Answer$\frac{\text{b}^2}{\sqrt{(\text{a}^2+\text{b}^2)}+\text{a}}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt{\text{a}^2+\text{b}^2}-\text{a}$
$=\frac{\text{b}^2\big(\sqrt{\text{a}^2+\text{b}^2}-\text{a}\big)}{\big(\sqrt{(\text{a}^2+\text{b}^2)}+\text{a}\big)\big(\sqrt{(\text{a}^2+\text{b}^2)}-\text{a}\big)}$
As we know, $(\text{a}-\text{b})^2=(\text{a}^2-2\times\text{a}\times\text{b}+\text{b}^2)$
$=\frac{\text{b}^2\big(\sqrt{\text{a}^2+\text{b}^2}-\text{a}\big)}{(\text{a}^2+\text{b}^2)-\text{a}^2}$
$=\frac{\text{b}^2\big(\sqrt{\text{a}^2+\text{b}^2}-\text{a}\big)}{\text{b}^2}$
View full question & answer→Question 53 Marks
Express each one of the following with rational denominator:$\frac{6-4\sqrt2}{6+4\sqrt{2}}$
Answer$\frac{6-4\sqrt2}{6+4\sqrt{2}}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $6-4\sqrt2$
$=\frac{(6-4\sqrt2)(6-4\sqrt2)}{(6+4\sqrt{2})(6-4\sqrt{2})}$
As we know, $(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$\frac{(6-4\sqrt2)^2}{36-32}$
As we know, $(\text{a}-\text{b})^2=(\text{a}^2-2\times\text{a}\times\text{b}+\text{b}^2)$
$=\frac{36-48\sqrt2+32}{4}$
$=\frac{68-48\sqrt2}{4}$
$=\frac{4(17-12\sqrt2)}{4}$
$=17-12\sqrt2$
View full question & answer→Question 63 Marks
In each of the following determine rational numbers a and b:$\frac{\sqrt3-1}{\sqrt3+1}=\text{a}-\text{b}\sqrt3$
AnswerWe know that rationalization factor for $\sqrt3+1$ is $\sqrt3-1.$ We will multiply numerator and denominator of the given expression $\frac{\sqrt3-1}{\sqrt3+1}$ by $\sqrt3-1,$ to get$\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1}=\frac{\big(\sqrt3\big)^2+(1)^2-2\times\sqrt3\times1}{\big(\sqrt3\big)^2-(1)^2}$
$=\frac{3+1-2\sqrt3}{3-1}$
$=\frac{4-2\sqrt3}{2}$
$=2-\sqrt3$
On equating rational and irrational terms, we get$\text{a}-\text{b}\sqrt3=2-\sqrt3$
$=2-1\sqrt3$
Hence, we get a = 2, b = 1.
View full question & answer→Question 73 Marks
Express each one of the following with rational denominator:$\frac{3\sqrt2+1}{2\sqrt{5}-3}$
Answer$\frac{3\sqrt2+1}{2\sqrt{5}-3}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $2\sqrt5-3$
$=\frac{(3\sqrt2+1)\times(2\sqrt5-3)}{(2\sqrt{5}-3)(2\sqrt{5}-3)}$
As we know, $(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{6\sqrt{10}-9\sqrt2+2\sqrt5-3}{(20-9)}$
$=\frac{6\sqrt{10}-9\sqrt2+2\sqrt5-3}{11}$
View full question & answer→Question 83 Marks
Simplify:$\frac{1}{2+\sqrt3}+\frac{2}{\sqrt5-\sqrt3}+\frac{1}{2-\sqrt5}$
AnswerWe know that rationalization factor for $2+\sqrt3,\ \sqrt5-\sqrt3,$ and $2-\sqrt5$ are $2-\sqrt3,\ \sqrt5+\sqrt3$ and $2+\sqrt5$ respectively. We will multiply numerator and denominator of the given expression $\frac{1}{2+\sqrt3},\ \frac{2}{\sqrt5-\sqrt3}$ and $\frac{1}{2-\sqrt5}$ by $2-\sqrt3.$$\sqrt5+\sqrt3$ and $2+\sqrt5$ respectively, to get
$\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}+\frac{2}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}+\frac{1}{2-\sqrt5}\times\frac{2+\sqrt5}{2+\sqrt5}\\ \ \ =\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}+\frac{2\sqrt5+2\sqrt3}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}+\frac{2-\sqrt5}{(2)^2-\big(\sqrt5\big)^2}$
$=\frac{2-\sqrt{3}}{1}+\frac{2\sqrt{5}+2\sqrt3}{5-3}+\frac{2+\sqrt5}{4-5}$
$=\frac{2-\sqrt{3}}{1}+\frac{2\sqrt{5}+2\sqrt3}{2}+\frac{2+\sqrt5}{-1}$
$=2-\sqrt3+\sqrt5+\sqrt3-\sqrt5-2$
$=0$
Hence the given expression is simplified to 0.
View full question & answer→Question 93 Marks
If $\text{x}=\sqrt2-1,$ then write the value of $\frac{1}{\text{x}}.$
AnswerGiven that $\text{x}=\sqrt2-1.$ Hence $\frac{1}{\text{x}}$ is given as$\frac{1}{\text{x}}=\frac{1}{\sqrt2-1}$
We know that rationalization factor for $\sqrt2-1$ is $\sqrt2+1.$ We will multiply each side of the given expression $\frac{1}{\sqrt2-1}$ by $\sqrt2+1,$ to get$\frac{1}{\sqrt2-1}\times\frac{\sqrt2+1}{\sqrt2+1}=\frac{\sqrt2+1}{\big(\sqrt2\big)^2-(1)^2}$
$=\frac{\sqrt2+1}{\sqrt2-1}$
$=\sqrt2+1$
Hence the value of the given expresion is $\sqrt2+1.$
View full question & answer→Question 103 Marks
In the following determine rational numbers a and b:$\frac{4+\sqrt2}{2+\sqrt2}=\text{a}-\sqrt{\text{b}}$
AnswerWe know that rationalization factor for $2+\sqrt2$ is $2-\sqrt2.$ We will multiply numerator and denominator of the given expression $\frac{4+\sqrt2}{2+\sqrt2}$ by $2-\sqrt2,$ to get$\frac{4+\sqrt2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2}=\frac{4\times2-4\times\sqrt2+2\times\sqrt2-\big(\sqrt2\big)^2}{(2)^2-\big(\sqrt2\big)^2}$
$=\frac{8-4\sqrt2+2\sqrt2-2}{4-2}$
$=\frac{6-2\sqrt2}{2}$
$=3-\sqrt2$
On equating rational and irrational terms, we get$\text{a}-\sqrt{\text{b}}=3-\sqrt2$
Hence, we get a = 3, b = 2.
View full question & answer→Question 113 Marks
Simplify:$\frac{2}{\sqrt5+\sqrt3}+\frac{1}{\sqrt3+\sqrt2}-\frac{3}{\sqrt5+\sqrt2}$
AnswerWe know that rationalization factor for $\sqrt5+\sqrt3,\ \sqrt3+\sqrt2,$ and $\sqrt5+\sqrt2$ are $\sqrt5+\sqrt3,\ \sqrt3-\sqrt2$ and $\sqrt5-\sqrt2$ respectively. We will multiply numerator and denominator of the given expression $\frac{2}{\sqrt5+\sqrt3},\ \frac{1}{\sqrt3+\sqrt2}$ and $\frac{3}{\sqrt5+\sqrt2}$ by $2-\sqrt3,$ $\sqrt5+\sqrt3$ and $2+\sqrt5$ respectively, to get$\frac{2}{\sqrt5+\sqrt2}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}+\frac{1}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}-\frac{3}{\sqrt5+\sqrt2}\times\frac{\sqrt5-\sqrt2}{\sqrt5-\sqrt2}\\ \ \ =\frac{2\sqrt5-2\sqrt3}{5-3}+\frac{\sqrt3-\sqrt2}{3-2}-\frac{3\sqrt5-3\sqrt2}{5-2}$
$=\frac{2\sqrt5-2\sqrt{3}}{2}+\frac{\sqrt{3}-\sqrt2}{1}-\frac{3\sqrt5-3\sqrt2}{3}$
$=\sqrt5-\sqrt3+\sqrt3-\sqrt2-\sqrt5+\sqrt2$
$=0$
Hence the given expression is simplified to 0.
View full question & answer→Question 123 Marks
In each of the following determine rational numbers a and b:$\frac{\sqrt{11}-\sqrt7}{\sqrt11+\sqrt7}=\text{a}-\text{b}\sqrt{77}$
AnswerWe know that rationalization factor for $\sqrt{11}+\sqrt7$ is $\sqrt{11}-\sqrt7.$ We will multiply numerator and denominator of the given expression $\frac{\sqrt{11}-\sqrt7}{\sqrt{11}-\sqrt7}$ by $\sqrt{11}-\sqrt7,$ to get$\frac{\sqrt{11}-\sqrt7}{\sqrt{11}-\sqrt7}\times\frac{\sqrt{11}-\sqrt7}{\sqrt{11}-\sqrt7}=\frac{\big(\sqrt{11}\big)^2+\big(\sqrt7\big)^2-2\times\sqrt{11}\times\sqrt7}{\big(\sqrt{11}\big)^2-\big(\sqrt7\big)^2}$
$=\frac{11+7-2\sqrt{77}}{11-7}$
$=\frac{18-2\sqrt{77}}{4}$
$=\frac{9}{2}-\frac{1}{2}\sqrt{77}$
On equating rational and irrational terms, we get$\text{a}+\text{b}\sqrt{77}=\frac{9}{2}-\frac{1}{2}\sqrt{77}$
Hence, we get $\text{a}=\frac{9}{2},\ \text{b}=\frac{1}{2}.$
View full question & answer→Question 133 Marks
Rationales the denominator and simplify:$\frac{2\sqrt6-\sqrt5}{3\sqrt5-2\sqrt6}$
Answer$\frac{2\sqrt6-\sqrt5}{3\sqrt5-2\sqrt6}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $3\sqrt5+2\sqrt6$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{\big(3\sqrt5-2\sqrt6\big)\big(3\sqrt5+2\sqrt6\big)}$
As we know, $(\text{a}-\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{45-24}$
$=\frac{\big(2\sqrt6-\sqrt5\big)\big(3\sqrt5+2\sqrt6\big)}{21}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{21}$
$=\frac{4\sqrt{30}+9}{21}$
View full question & answer→Question 143 Marks
Express each one of the following with rational denominator:$\frac{30}{5\sqrt3-3\sqrt5}$
Answer$\frac{30}{5\sqrt3-3\sqrt5}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $5\sqrt3+3\sqrt5$
$=\frac{30\times(5\sqrt3+3\sqrt5)}{(5\sqrt3-3\sqrt5)(5\sqrt3+3\sqrt5)}$
As we know $\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2$
$=\frac{30\times(5\sqrt3+3\sqrt5)}{75-45}$
$=\frac{30\times(5\sqrt3+3\sqrt5)}{30}$
$=5\sqrt3+3\sqrt5$
View full question & answer→Question 153 Marks
Rationales the denominator and simplify:$\frac{5+2\sqrt3}{7+4\sqrt3}$
Answer$\frac{5+2\sqrt3}{7+4\sqrt3}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $7-4\sqrt3$
$=\frac{\big(5+2\sqrt3\big)\big(\sqrt3-\sqrt2\big)}{\big(\sqrt3+\sqrt2\big)\big(\sqrt3-\sqrt2\big)}$
As we know, $(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(5+2\sqrt3\big)^2\big(7-4\sqrt3\big)}{49-48}$
$=35-20\sqrt3+14\sqrt3-24=11-6\sqrt3$
View full question & answer→Question 163 Marks
In each of the following determine rational numbers a and b:$\frac{5+3\sqrt3}{7+4\sqrt3}=\text{a}+\text{b}\sqrt3$
AnswerWe know that rationalization factor for $7+4\sqrt3$ is $7-4\sqrt3.$ We will multiply numerator and denominator of the given expression $\frac{5+3\sqrt3}{7+4\sqrt3}$ by $7-4\sqrt3$ to get$\frac{5+3\sqrt3}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3}=\frac{5\times7-5\times4\times\sqrt3+3\times7\times\sqrt3-3\times4\times\big(\sqrt3\big)^2}{(7)^2-\big(4\sqrt3\big)^2}$
$=\frac{35-20\sqrt3+21\sqrt3-36}{49-48}$
$=\frac{\sqrt3-1}{1}$
$=\sqrt3-1$
On equating rational and irrational terms, we get$\text{a}+\text{b}\sqrt3=\sqrt3-1$
$=-1+1\sqrt3$
Hence, we get a = -1, b = 1.
View full question & answer→Question 173 Marks
Rationales the denominator and simplify:$\frac{2\sqrt3-\sqrt5}{2\sqrt{2}+3\sqrt{3}}$
Answer$\frac{2\sqrt3-\sqrt5}{2\sqrt{2}+3\sqrt{3}}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $2\sqrt{2}-3\sqrt{3}$
$=\frac{\big(2\sqrt3+\sqrt5\big)\big(2\sqrt{2}-3\sqrt{3}\big)}{\big(2\sqrt{2}+3\sqrt{3}\big)\big(2\sqrt{2}-3\sqrt{3}\big)}$
$=\frac{\big(2\sqrt3+\sqrt5\big)\big(2\sqrt{2}-3\sqrt{3}\big)}{8-27}$
$=\frac{\big(4\sqrt6-2\sqrt{10}\big)-18+3\sqrt{15}}{30}$
$=\frac{18-4\sqrt6+2\sqrt{10}-3\sqrt{15}}{19}$
View full question & answer→Question 183 Marks
In the following determine rational numbers a and b:$\frac{3+\sqrt2}{3-\sqrt2}=\text{a}+\text{b}\sqrt{2}$
AnswerWe know that rationalization factor for $3-\sqrt2$ is $3+\sqrt2.$ We will multiply numerator and denominator of the given expression $\frac{3+\sqrt2}{3-\sqrt2}$ by $3+\sqrt2,$ to get$\frac{3+\sqrt2}{3-\sqrt2}\times\frac{3+\sqrt2}{3+\sqrt2}=\frac{(3)^2+\big(\sqrt2\big)^2+2\times3\times\sqrt2}{(3)^2-\big(\sqrt2\big)^2}$
$=\frac{9+2+6\sqrt2}{9-2}$
$=\frac{11+6\sqrt2}{7}$
$=\frac{11}{7}+\frac{6}{7}\sqrt2$
On equating rational and irrational terms, we get$\text{a}+\text{b}\sqrt2=\frac{11}{6}+\frac{6}{7}\sqrt2$
Hence, we get $\text{a}=\frac{11}{7},\ \text{b}=\frac{6}{7}.$
View full question & answer→Question 193 Marks
Express each one of the following with rational denominator:$\frac{16}{\sqrt{41}-5}$
Answer$\frac{16}{\sqrt{41}-5}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt{41}+5$
$=\frac{16\times(\sqrt{41}+5)}{(\sqrt{41}-5)(\sqrt{41}+5)}$
As we know,
$(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)=\frac{16\sqrt{41}}{41-5}$
$=\frac{16\sqrt{41}+80}{16}$
$=\frac{16(\sqrt{41}+5)}{16}$
$=\sqrt{41}+5$
View full question & answer→Question 203 Marks
Rationales the denominator and simplify:$\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}$
Answer$\frac{4\sqrt3+5\sqrt2}{\sqrt{48}+\sqrt{18}}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $\sqrt{48}-\sqrt{18}$
$=\frac{\big(4\sqrt3+5\sqrt2\big)\big(\sqrt{48}-\sqrt{18}\big)}{\big(\sqrt{48}+\sqrt{18}\big)\big(\sqrt{48}-\sqrt{18}\big)}$
As we know, $(\text{a}+\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(4\sqrt3+5\sqrt2\big)\big(\sqrt{48}-\sqrt{18}\big)}{48-18}$
$=\frac{48-12\sqrt6+20\sqrt6-30}{30}$
$=\frac{18+8\sqrt6}{30}$
$=\frac{9+4\sqrt6}{15}$
View full question & answer→Question 213 Marks
Simplify:$\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}+\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$
AnswerWe know that rationalization factor for $\sqrt5-\sqrt3$ and $\sqrt5+\sqrt3$ are $\sqrt5+\sqrt3$ and $\sqrt5-\sqrt3$ respectively. We will multiply numerator and denominator of the given expression $\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$ and $\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$ by $\sqrt5+\sqrt3$ and $\sqrt5+\sqrt3$ respectively, to get$\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}+\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}\\=\frac{\big(\sqrt5\big)^2+\big(\sqrt3\big)^2+2\times\sqrt5\times\sqrt3}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}+\frac{\big(\sqrt5\big)^2+\big(\sqrt3\big)^2-2\times\sqrt5\times\sqrt3}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}$
$=\frac{5+3+2\sqrt{15}}{5-3}+\frac{5+3-2\sqrt{15}}{5-3}$
$=\frac{5+3+2\sqrt{15}+5+3-2\sqrt{15}}{2}$
$=\frac{16}{2}$
$=8$
Hence the given expression is simplified to 8.
View full question & answer→Question 223 Marks
Rationales the denominator and simplify:$\frac{1+\sqrt2}{3-2\sqrt2}$
Answer$\frac{1+\sqrt2}{3-2\sqrt2}$Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor $3+2\sqrt2$
$=\frac{\big(1+\sqrt2\big)\big(3+2\sqrt2\big)}{\big(3-2\sqrt2\big)\big(3+2\sqrt2\big)}$
As we know, $(\text{a}-\text{b})(\text{a}-\text{b})=(\text{a}^2-\text{b}^2)$
$=\frac{\big(1+\sqrt2\big)\big(3+2\sqrt2\big)}{9-8}$
$=3+2\sqrt2+3\sqrt2+4=7+5\sqrt2$
View full question & answer→Question 233 Marks
In each of the following determine rational numbers a and b:$\frac{4+3\sqrt5}{4-3\sqrt5}=\text{a}+\text{b}\sqrt{5}$
AnswerWe know that rationalization factor for $4+3\sqrt5$ is $4+3\sqrt5.$ We will multiply numerator and denominator of the given expression $\frac{4+3\sqrt5}{4-3\sqrt5}$ by $4+3\sqrt5,$ to get$\frac{4+3\sqrt5}{4-3\sqrt5}\times\frac{4+3\sqrt5}{4+3\sqrt5}=\frac{(4)^2+\big(3\sqrt5\big)^2+2\times4\times3\sqrt5}{(4)^2-\big(3\sqrt5\big)^2}$
$=\frac{16+45+24\sqrt{5}}{16-45}$
$=\frac{61+24\sqrt{5}}{-29}$
$=-\frac{61}{29}-\frac{24}{29}\sqrt{5}$
On equating rational and irrational terms, we get$\text{a}+\text{b}\sqrt{5}=-\frac{61}{29}-\frac{24}{29}\sqrt{5}$
Hence, we get $\text{a}=-\frac{61}{29},\ \text{b}=-\frac{24}{29}.$
View full question & answer→Question 243 Marks
Simplify:$\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}+\frac{\sqrt{12}}{\sqrt3-\sqrt2}$
AnswerWe know that rationalization factor for $3\sqrt2+2\sqrt3$ and $\sqrt3-\sqrt2$ are $3\sqrt2-2\sqrt3$ and $\sqrt3+\sqrt2$ respectively. We will multiply numerator and denominator of the given expression $\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}$ and $\frac{\sqrt{12}}{\sqrt3-\sqrt2}$ by $3\sqrt2-2\sqrt3,$ and $\sqrt3+\sqrt2$ respectively. to get$\frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}\times\frac{3\sqrt2-2\sqrt3}{3\sqrt2-2\sqrt3}+\frac{\sqrt{12}}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ \ =\frac{\big(3\sqrt2\big)^2+\big(2\sqrt3\big)^2-2\times3\sqrt2\times2\sqrt3}{\big(3\sqrt2\big)^2-\big(2\sqrt3\big)^2}+\frac{\sqrt{36}+\sqrt{24}}{\big(\sqrt3\big)^2-\big(\sqrt2\big)^2}$
$=\frac{18+12-12\sqrt6}{18-12}+\frac{6+\sqrt{24}}{3-2}$
$=\frac{30-12\sqrt6+36+12\sqrt6}{6}$
$=\frac{66}{6}$
$=11$
Hence the given expression is simplified to 11.
View full question & answer→