If a2-1-a2-a4-1-a4 then find the value of a-1-a

The correct option is D none of thesea^2 + 1/a^2 = a^4 + 1/a^4 a^4 + 1/a^2 = a^8 + 1/a^4 Or, (a^4 + 1) a^2 = a^8 + 1 Or, a^6 + a^2 a^8 1 = 0 Or, a^6 (1 ...

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Byju's Answer

Standard IX

Mathematics

Laws of Exponents for Real Numbers

If a2+1/a2=a4...

Question

If $a^{2} + \frac{1}{a^{2}} = a^{4} + \frac{1}{a^{4}}$ then find the value of $a + \frac{1}{a}$

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none of these

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Solution

The correct option is D
none of these

$a^{2} + \frac{1}{a^{2}} = a^{4} + \frac{1}{a^{4}} \frac{a^{4} + 1}{a^{2}} = \frac{a^{8} + 1}{a^{4}} O r, (a^{4} + 1) a^{2} = a^{8} + 1 O r, a^{6} + a^{2} - a^{8} - 1 = 0 O r, a^{6} (1 - a^{2}) - 1 (1 - a^{2}) = 0 O r, (1 - a^{2}) (a^{6} - 1) = 0$
After solving this we get $a = \pm 1$
So, the value of $a + \frac{1}{a} = \pm 2$

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If a2+1a2=a4+1a4 then find the value of a+1a

The correct option is D none of thesea2+1a2=a4+1a4a4+1a2=a8+1a4Or,(a4+1)a2=a8+1Or,a6+a2−a8−1=0Or,a6(1−a2)−1(1−a2)=0Or,(1−a2)(a6−1)=0 After solving this we get a=±1 So, the value of a+1a=±2

If $a^{2} + \frac{1}{a^{2}} = a^{4} + \frac{1}{a^{4}}$ then find the value of $a + \frac{1}{a}$