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Maths 4 Mark Question

Rationalisation · Maharashtra Board STD 9 (MSBSHSE - English Medium). 9 questions.

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4 Mark Question

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Question 14 Marks
If $\text{a}=\sqrt2+1,$ then write the value of $\text{a}-\frac{1}{\text{a}}.$
Answer
Given that $\text{a}=\sqrt2+1,$ Hence $\frac{1}{\text{a}}$ is given as$\frac{1}{\text{a}}=\frac{1}{\sqrt2+1}.$ we are asked to find $\text{a}-\frac{1}{\text{a}}$
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$\frac{1}{\text{a}}=\frac{1}{\sqrt2-1}\times\frac{\sqrt2-1}{\sqrt2-1}$
$=\frac{\sqrt2-1}{\big(\sqrt2\big)^2-(1)^2}$
$=\frac{\sqrt2-1}{\sqrt2-1}$
$=\sqrt2-1$
Therefore,$\text{a}-\frac{1}{\text{a}}=\sqrt2+1-\big(\sqrt2-1\big)$
$=\sqrt2+1-\sqrt2+1$
$=2$
Hence the value of the given expresion is 2.
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Question 24 Marks
Find the value of $\frac{6}{\sqrt5-\sqrt3},$ it being given that $\sqrt3=1.732$ and $\sqrt5=2.236.$
Answer
We know that rationalization factor for $\sqrt5-\sqrt3$ is $\sqrt5+\sqrt3.$ We will multiply numerator and denominator of the given expression $\frac{6}{\sqrt5-\sqrt3}$ by $\sqrt5+\sqrt3,$ to get$\frac{6}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}=\frac{6\sqrt5+6\sqrt3}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}$
$=\frac{6\sqrt5+6\sqrt{3}}{5-3}$
$=\frac{6\sqrt{5}+6\sqrt3}{2}$
$=3\sqrt5+3\sqrt3$
Putting the value of $\sqrt5$ and $\sqrt3,$ we get$3\sqrt5+3\sqrt3=3(2.236)+3(1.732)$
$=6.708+5.196$
$=11.904$
Hence value of the given expression is 11.904.
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Question 34 Marks
Find the value of the following correct to three place of decimals, it begin that $\sqrt2=1.4142, \sqrt3=1.732,\ \sqrt5=2.2360,\ \sqrt6=2.4495$ and $\sqrt{10}=3.162.$$\frac{3-\sqrt5}{3+2\sqrt5}$
Answer
We know that rationalization factor for $3+2\sqrt5$ is $3-2\sqrt5.$ We will multiply numerator and denominator of the given expression $\frac{3-\sqrt5}{3+2\sqrt5}$ by $3-2\sqrt5,$ to get$\frac{3-\sqrt5}{3+2\sqrt5}\times\frac{3-2\sqrt5}{3-2\sqrt5}=\frac{(3)^2-3\times2\times\sqrt5-3\times\sqrt5+2\times\big(\sqrt5\big)^2}{(3)^2-\big(2\sqrt5\big)^2}$
$=\frac{9-9\sqrt5+10}{9-20}$
$=\frac{19-9\sqrt5}{-11}$
$=\frac{9\sqrt5-19}{11}$
Putting the value of $\sqrt5,$ we get$\frac{9\sqrt5-19}{11}=\frac{9(2.236)-19}{11}$
$=\frac{20.124-19}{11}$
$=\frac{1.124}{11}$
$=0.102$
Hence the given expression is simplified to 0.102.
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Question 44 Marks
Simplify:$\frac{7+3\sqrt5}{3+\sqrt5}-\frac{7-3\sqrt5}{3-\sqrt5}$
Answer
We know that rationalization factor for $3+\sqrt5$ and $3-\sqrt5$ are $3-\sqrt5$ and $3+\sqrt5$ respectively. We will multiply numerator and denominator of the given expression $\frac{7+3\sqrt5}{3+\sqrt5}$ and $\frac{7-3\sqrt5}{3+\sqrt5}$ by $3-\sqrt5$ and $3+\sqrt5$ respectively. to get$\frac{7+3\sqrt5}{3+\sqrt5}\times\frac{3-\sqrt5}{3-\sqrt5}-\frac{7-3\sqrt5}{3-\sqrt5}\times\frac{3+\sqrt5}{3+\sqrt5}\\ \ =\frac{7\times3-7\times\sqrt5+9\times\sqrt5-3\times\big(\sqrt5\big)^2}{(3)^2-\big(\sqrt5\big)^2}-\frac{7\times3+7\times\sqrt5-9\times\sqrt5-3\times\big(\sqrt5\big)^2}{(3)^2-\big(\sqrt5\big)^2}$
$=\frac{21-7\sqrt5+9\sqrt5-3\times5}{9-5}-\frac{21+7\sqrt5-9\sqrt5-3\times5}{9-5}$
$=\frac{21+2\sqrt5-15}{4}-\frac{21-2\sqrt5-15}{4}$
$=\frac{6+2\sqrt5-6+2\sqrt5}{4}$
$=\frac{4\sqrt5}{4}$
$=\sqrt5$
Hence the given expression is simplified to $\sqrt5.$
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Question 54 Marks
If $\text{x}=2+\sqrt3,$ find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$
Answer
We know that $\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2-1+\frac{1}{\text{x}^2}\Big).$ we have to find the value of $\text{x}^3+\frac{1}{\text{x}^3}.$ As $\text{x}=2+\sqrt3$ therefore,$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}$
We know that rationalization factor for $2+\sqrt3$ is $2-\sqrt3.$ We will multiply numerator and denominator of the given expression $\frac{1}{2+\sqrt3}$ by $2-\sqrt3,$ to get.$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{2-\sqrt3}{4-3}$
$=2-\sqrt3$
Putting the value of x and $\frac{1}{\text{x}},$ we get$\text{x}^3+\frac{1}{\text{x}^3}=\big(2+\sqrt3+2-\sqrt3\big)\Big(\big(2+\sqrt3\big)^2-1+\big(2-\sqrt3\big)^2\Big)$
$=4\Big(2^2+\big(\sqrt3\big)^2+2\times2\times\sqrt3-1+2^2+\big(\sqrt3\big)^2-2\times2\times\sqrt3\Big)$
$=4\big(4+3+4\sqrt3-1+4+3-4\sqrt3\big)$
$=52$
Hence the value of the given expression 52.
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Question 64 Marks
Find the value of the following correct to three place of decimals, it begin that $\sqrt2=1.4142, \sqrt3=1.732,\ \sqrt5=2.2360,\ \sqrt6=2.4495$ and $\sqrt{10}=3.162.$$\frac{1+\sqrt2}{3-2\sqrt2}$
Answer
We know that rationalization factor for $3-2\sqrt5$ is $3+2\sqrt5.$ We will multiply numerator and denominator of the given expression $\frac{1+\sqrt2}{3-2\sqrt2}$ by $3+2\sqrt5,$ to get$\frac{1+\sqrt2}{3-2\sqrt2}\times\frac{3+2\sqrt2}{3+2\sqrt2}=\frac{3+2\times\sqrt2+3\times\sqrt2+2\times\big(\sqrt2\big)^2}{(3)^2-\big(2\sqrt2\big)^2}$
$=\frac{3+2\sqrt2+3\sqrt2+4}{9-8}$
$=\frac{7+5\sqrt2}{1}$
$=7+5\sqrt2$
Putting the value of $\sqrt2,$ we get$7+5\sqrt2=7+5(1.4142)$
$=7+7.071$
$=14.071$
Hence the given expression is simplified to 14.071.
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Question 74 Marks
If $\text{x}=3+\sqrt8,$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
Answer
We know that $\text{x}^2+\frac{1}{\text{x}^2}=\Big(\text{x}+\frac{1}{\text{x}^2}\Big)^2-2.$ we have to find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$ As $\text{x}=3+\sqrt8$ therefore,$\frac{1}{\text{x}}=\frac{1}{3+\sqrt8}$
We know that rationalization factor for $3+\sqrt8$ is $3-\sqrt8.$ We will multiply numerator and denominator of the given expression $\frac{1}{3+\sqrt8}$ by $3-\sqrt8,$ to get.$\frac{1}{\text{x}}=\frac{1}{3+\sqrt8}\times\frac{3-\sqrt8}{3-\sqrt8}$
$=\frac{3-\sqrt8}{(3)^2-\big(\sqrt8\big)^2}$
$=\frac{3-\sqrt8}{9-8}$
$=3-\sqrt8$
Putting the value of x and $\frac{1}{\text{x}},$ we get$\text{x}^2+\frac{1}{\text{x}^2}=\big(3+\sqrt8+3-\sqrt8\big)^2-2$
$=(6)^2-2$
$=36-2$
$=34$
Hence the given expression is simplified to 34.
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Question 84 Marks
If $\text{x}=2+\sqrt3,$ find the value of $\text{x}+\frac{1}{\text{x}}.$
Answer
Given that $\text{x}=2+\sqrt3,$ Hence $\frac{1}{\text{x}}$ is given as$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}.$ we are asked to find $\text{x}+\frac{1}{\text{x}}$
We know that rationalization factor for $2+\sqrt3$ is $2-\sqrt3.$ We will multiply each side of the given expression $\frac{1}{2+\sqrt3}$ by $2-\sqrt3,$ to get$\frac{1}{\text{x}}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{2-\sqrt3}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{2-\sqrt3}{4-3}$
$=2-\sqrt3$
Therefore,$\text{x}+\frac{1}{\text{x}}=2+\sqrt3+2-\sqrt3$
$=4$
Hence the value of the given expresion is 4.
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Question 94 Marks
If $\text{x}=\frac{\sqrt3+1}{2},$ find the value of $4\text{x}^3+2\text{x}^2-8\text{x}+7.$
Answer
We have, $\text{x}=\frac{\sqrt3+1}{2}$ It can be simplified as$2\text{x}-1=\sqrt3$
On squaring both sides, we get$(2\text{x}-1)^2=\big(\sqrt3\big)^2$
$(2\text{x})^2+1-2\times2\text{x}=3$
$4\text{x}^2+1-4\text{x}=3$
$4\text{x}^2-4\text{x}-2=0$
The given equation can be rewritten as. $4\text{x}^3+2\text{x}^2-8\text{x}+7\\ \ =\text{x}(4\text{x}^2-4\text{x}-2)+\frac{6}{4}(4\text{x}^2-4\text{x}-2)+3+7$
Therefore, we have$4\text{x}^3+2\text{x}^2-8\text{x}+7=\text{x}(0)+\frac{6}{4}(0)+3+7$
$=3+7$
$=10$
Hence, the value of given expression is 10.
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