Question

if a+1/a=6 find a- 1/a and a^2- 1/a^2

Answer 1

Hi...☺

Here is your answer...✌

GIVEN THAT,

$a + \frac{1}{a} = 6 \\ \\ Squaring \: both \: sides \\ we \:get \\ \\ {(a + \frac{1}{a} ) }^{2} = {6}^{2} \\ \\ {a}^{2} + { \frac{1}{a {}^{2} } } + 2 \times a \times \frac{1}{a} = 36 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 36 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } = 34 \\ \\ Subtracting \: \: 2\: \: both \: \: side s \\ we \: get \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } - 2 = 34 - 2 \\ \\ {a}^{2} + \frac{1}{ {a}^{2} } - 2 \times a \times \frac{1}{a} = 32 \\ \\ {(a - \frac{1}{a} ) }^{2} = 32 \\ \\ {(a - \frac{1}{a} ) }= \sqrt{32} \:\: or \: \: -\sqrt{32} \\ \\ {(a - \frac{1}{a} ) }= 4\sqrt{2} \: \: or \: \: - 4 \sqrt{2} \\ \\Now, \\\\ {a}^{2} - \frac{1}{ {a}^{2} } = (a + \frac{1}{a} )(a - \frac{1}{a} ) \\ \\ when \: (a - \frac{1}{a} ) = 4 \sqrt{2 } \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 6 \times 4 \sqrt{2} \\ \\ = 24 \sqrt{2} \\ \\ when \: (a - \frac{1}{a} ) = - 4 \sqrt{2} \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 6 \times ( - 4 \sqrt{2} ) \\ \\ = - 24 \sqrt{2}$

HENCE,

$(a - \frac{1}{a} ) = 4 \sqrt{2} \: \: or \: \: - 4 \sqrt{2} \\ \\ and \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 24 \sqrt{2} \: \: or \: \: - 24 \sqrt{2}$

Answer 2

Step-by-step explanation:

a+a1=6Squaringbothsidesweget(a+a1)2=62a2+a21+2×a×a1=36a2+a21+2=36a2+a21=34Subtracting2bothsideswegeta2+a21−2=34−2a2+a21−2×a×a1=32(a−a1)2=32(a−a1)=32or−32(a−a1)=42or−42Now,a2−a21=(a+a1)(a−a1)when(a−a1)=42(a2−a21)=6×42=242when(a−a1)=−42(a2−a21)=6×(−42)=−242

HENCE,

\begin{gathered}(a - \frac{1}{a} ) = 4 \sqrt{2} \: \: or \: \: - 4 \sqrt{2} \\ \\ and \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 24 \sqrt{2} \: \: or \: \: - 24 \sqrt{2} \end{gathered}(a−a1)=42or−42and(a2−a21)=242or−242

answer