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Mathematics

If a2−4a−1=0a^2 - 4a - 1 = 0 and a≠0a \neq 0, find the values of:

(i) (a−1a)\Big(a - \dfrac{1}{a}\Big)

(ii) (a+1a)\Big(a + \dfrac{1}{a}\Big)

(iii) (a2−1a2)\Big(a^2 - \dfrac{1}{a^2}\Big)

(iv) (a2+1a2)\Big(a^2 + \dfrac{1}{a^2}\Big)

Expansions

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Answer

(i) Solving,

⇒a2−4a−1=0⇒a2−4a=1⇒a(a−4)=1⇒a−4=1a⇒a−1a=4\Rightarrow a^2 - 4a - 1 = 0 \\[1em] \Rightarrow a^2 - 4a = 1 \\[1em] \Rightarrow a(a - 4) = 1 \\[1em] \Rightarrow a - 4 = \dfrac{1}{a} \\[1em] \Rightarrow a - \dfrac{1}{a} = 4 \\[1em]

Hence, (a−1a)=4\Big(a - \dfrac{1}{a}\Big) = 4

(ii) Using identity,

⇒(a+1a)2−(a−1a)2=4⇒(a+1a)2−(4)2=4⇒(a+1a)2−16=4⇒(a+1a)2=4+16⇒(a+1a)2=20⇒a+1a=20⇒a+1a=±25.\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - (4)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 16 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 4 + 16\\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 20 \\[1em] \Rightarrow a + \dfrac{1}{a} = \sqrt{20} \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}.

Hence, a+1a=±25a + \dfrac{1}{a} = \pm 2\sqrt{5}

(iii) From part (i) and (ii),

⇒a−1a=4⇒a+1a=±25\Rightarrow a - \dfrac{1}{a} = 4 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}

Case 1:

⇒a+1a=25\Rightarrow a + \dfrac{1}{a} = 2\sqrt{5}

Using identity,

⇒(a+1a)(a−1a)=(a2−1a2)⇒(25)×(4)=(a2−1a2)⇒(a2−1a2)=85\Rightarrow \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (2\sqrt{5}) \times (4) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \dfrac{1}{a^2}\Big) = 8\sqrt{5} \\[1em]

Case 2:

⇒a+1a=−25\Rightarrow a + \dfrac{1}{a} = -2\sqrt{5}

Using identity,

⇒(a+1a)(a−1a)=(a2−1a2)⇒(−25)×(4)=(a2−1a2)⇒(a2−1a2)=−85\Rightarrow \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (-2\sqrt{5}) \times (4) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \dfrac{1}{a^2}\Big) = -8\sqrt{5} \\[1em]

Hence, (a2−1a2)=±85\Big( a^2 - \dfrac{1}{a^2}\Big) = \pm 8\sqrt{5}.

(iv) From part (i) and (ii),

⇒a−1a=4⇒a+1a=±25\Rightarrow a - \dfrac{1}{a} = 4 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}

Using identity,

⇒(a+1a)2+(a−1a)2=2(a2+1a2)⇒(±25)2+(4)2=2(a2+1a2)⇒2(a2+1a2)=20+16⇒2(a2+1a2)=36⇒(a2+1a2)=362⇒(a2+1a2)=18\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 + \Big(a - \dfrac{1}{a}\Big)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (\pm 2\sqrt{5})^2 + (4)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 20 + 16 \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 36 \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = \dfrac{36}{2} \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = 18 \\[1em]

Hence, (a2+1a2)=18\Big(a^2 + \dfrac{1}{a^2}\Big) = 18.

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