If x Find the value of I ii iii iv

Selina Solutions Class 9 Mathematics Solutions for Exercise 1(D) in Chapter 1 - Chapter 1- Rational and Irrational Numbers

Question 4 Exercise 1(D)

If x =√𝟑 − √𝟐. Find the value of:

(I) $x+\frac{1}{x}$

(ii) $x^{2}+\frac{1}{x^{2}}$

(iii) $x^{3}+\frac{1}{x^{3}}$

(iv) $x^{3}+\frac{1}{x^{3}}-3\left(x^{2}+\frac{1}{x^{2}}\right)+x+\frac{1}{x}$

Answer:

$\begin{aligned} &\text { (i) }\\ &x+\frac{1}{x} \end{aligned}$

$\begin{array}{l} \quad(\sqrt{3}-\sqrt{2})+\frac{1}{(\sqrt{3}-\sqrt{2})} \\ =\frac{(\sqrt{3}-\sqrt{2})^{2}+1}{(\sqrt{3}-\sqrt{2})} \\ =\frac{3-2 \sqrt{3} \sqrt{2}+2+1}{(\sqrt{3}-\sqrt{2})} \\ =\frac{6-2 \sqrt{3} \sqrt{2}}{(\sqrt{3}-\sqrt{2})} \\ =\frac{6-2 \sqrt{6}}{(\sqrt{3}-\sqrt{2})} \times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})} \\ -\frac{6 \sqrt{3}-2 \sqrt{6} \sqrt{3}+6 \sqrt{2}-2 \sqrt{6} \sqrt{2}}{6 \sqrt{3}-2 \sqrt{18}+6 \sqrt{2}-2 \sqrt{12}} \\ =6 \sqrt{3}-2 \sqrt{9 \times 2}+6 \sqrt{2}-2 \sqrt{4 \times 3} \\ =6 \sqrt{3}-2 \times 3 \sqrt{2}+6 \sqrt{2}-2 \times 2 \sqrt{3} \\ =6 \sqrt{3}-6 \sqrt{2}+6 \sqrt{2}-4 \sqrt{3} \\ =6 \sqrt{3}-4 \sqrt{3} \\ =2 \sqrt{3} \end{array}$

$\begin{aligned} &\text { (ii) }\\ &x^{2}+\frac{1}{x^{2}} \end{aligned}$

$\begin{array}{l} \quad(\sqrt{3}-\sqrt{2})^{2}+\frac{1}{(\sqrt{3}-\sqrt{2})^{2}} \\ -(3-2 \sqrt{3} \sqrt{2}+2)+\frac{1}{(3-2 \sqrt{3} \sqrt{2}+2)} \\ =(5-2 \sqrt{6})+\frac{1}{(5-2 \sqrt{6})} \\ =\frac{25-10 \sqrt{6}-10 \sqrt{6}+4 \times 6+1}{(5-2 \sqrt{6})} \\ -\frac{25-20 \sqrt{6}+25}{(5-2 \sqrt{6})} \\ =\frac{50-20 \sqrt{6}}{(5-2 \sqrt{6})} \\ -\frac{10(5-2 \sqrt{6})}{(5-2 \sqrt{6})} \\ =10 \end{array}$

$\begin{aligned} &\text { (iii) }\\ &x^{3}+\frac{1}{x^{3}} \end{aligned}$

$\begin{array}{l} \quad(\sqrt{3}-\sqrt{2})^{3}+\frac{1}{(\sqrt{3}-\sqrt{2})^{3}} \\ {\left[\begin{array}{l} \text { We know that }(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b) \\ (\sqrt{3}-\sqrt{2})^{3}=3 \sqrt{3}-2 \sqrt{2}-3 \sqrt{3} \sqrt{2}(\sqrt{3}-\sqrt{2}) \\ =3 \sqrt{3}-2 \sqrt{2}-3 \sqrt{6}(\sqrt{3}-\sqrt{2}) \\ =3 \sqrt{3}-2 \sqrt{2}-3 \sqrt{18}+3 \sqrt{12} \\ =3 \sqrt{3}-2 \sqrt{2}-3 \sqrt{9 \times 2}+3 \sqrt{4} \times 3 \\ =3 \sqrt{3}-2 \sqrt{2}-3 \times 3 \sqrt{2}+3 \times 2 \sqrt{3} \\ =3 \sqrt{3}-2 \sqrt{2}-9 \sqrt{2}+6 \sqrt{3} \\ =9 \sqrt{3}-11 \sqrt{2} \\ \therefore(\sqrt{3}-\sqrt{2})^{3}+\frac{1}{(\sqrt{3}-\sqrt{2})^{3}}=(9 \sqrt{3}-11 \sqrt{2})+\frac{1}{(9 \sqrt{3}-11 \sqrt{2})} \\ \text { Considering } \frac{1}{(9 \sqrt{3}-11 \sqrt{2})} \\ \frac{1}{(9 \sqrt{3}-11 \sqrt{2})} \times \frac{(9 \sqrt{3}+11 \sqrt{2})}{(9 \sqrt{3}+11 \sqrt{2})} \\ =\frac{(9 \sqrt{3}+11 \sqrt{2})}{(81 \times 3)-(121 \times 2)} \\ =\frac{(9 \sqrt{3}+11 \sqrt{2})}{(243)-(242)} \\ =(9 \sqrt{3}+11 \sqrt{2}) \\ \text { Now, }(9 \sqrt{3}-11 \sqrt{2})+\frac{1}{(9 \sqrt{3}-11 \sqrt{2})}=(9 \sqrt{3}-11 \sqrt{2})+(9 \sqrt{3}+11 \sqrt{2}) \\ =9 \sqrt{3}-11 \sqrt{2}+9 \sqrt{3}+11 \sqrt{2} \\ =18 \sqrt{3} \end{array}\right.} \end{array}$

$\begin{aligned} &\text { (iv) }\\ &x^{3}+\frac{1}{x^{3}}-3\left(x^{2}+\frac{1}{x^{2}}\right)+x+\frac{1}{x} \end{aligned}$

$\begin{aligned} &x^{3}+\frac{1}{x^{3}}-3\left(x^{2}+\frac{1}{x^{2}}\right)+x+\frac{1}{x}\\ &\text { According to the solutions obtained in }(i),(i i) \text { and }(i i i), \text { we get, }\\ &\begin{array}{l} x^{3}+\frac{1}{x^{3}}-3\left(x^{2}+\frac{1}{x^{2}}\right)+x+\frac{1}{x}=18 \sqrt{3}-3(10)+2 \sqrt{3} \\ =20 \sqrt{3}-30 \\ =10(2 \sqrt{3}-3) \end{array} \end{aligned}$

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