If a2−3a−1=0a^2 -3a -1 = 0a2−3a−1=0, find the value of a2+1a2a^2+\dfrac{1}{a^2}a2+a21 .
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Given,
a2−3a−1=0⇒a2−1=3a⇒a2−1a=3⇒a2a−1a=3⇒a−1a=3a^2 -3a -1 = 0\\[1em] ⇒ a^2 - 1 = 3a\\[1em] ⇒ \dfrac{a^2 - 1}{a} = 3\\[1em] ⇒ \dfrac{a^2}{a} - \dfrac{1}{a} = 3\\[1em] ⇒ a - \dfrac{1}{a} = 3\\[1em]a2−3a−1=0⇒a2−1=3a⇒aa2−1=3⇒aa2−a1=3⇒a−a1=3
Squaring both sides, we get:
⇒(a−1a)2=32⇒a2+1a2−2×a×1a=9⇒a2+1a2−2×a×1a=9⇒a2+1a2−2=9⇒a2+1a2=9+2⇒a2+1a2=11⇒ \Big(a - \dfrac{1}{a}\Big)^2 = 3^2\\[1em] ⇒ a^2 + \dfrac{1}{a}^2 - 2 \times a \times \dfrac{1}{a} = 9\\[1em] ⇒ a^2 + \dfrac{1}{a}^2 - 2 \times \cancel{a} \times \dfrac{1}{\cancel{a}} = 9\\[1em] ⇒ a^2 + \dfrac{1}{a}^2 - 2 = 9\\[1em] ⇒ a^2 + \dfrac{1}{a}^2 = 9 + 2\\[1em] ⇒ a^2 + \dfrac{1}{a}^2 = 11⇒(a−a1)2=32⇒a2+a12−2×a×a1=9⇒a2+a12−2×a×a1=9⇒a2+a12−2=9⇒a2+a12=9+2⇒a2+a12=11
Hence, a2+1a2a^2+\dfrac{1}{a^2}a2+a21 = 11.
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If a + b = 7 and ab = 6, find a2−b2a^2 - b^2a2−b2.
If a2+b2=13a^2 + b^2 = 13a2+b2=13 and ab = 6, find :
(i) a + b
(ii) a - b
(iii) a2−b2a^2 - b^2a2−b2
(iv) 3(a+b)2−2(a−b)23(a+b)^2-2(a-b)^23(a+b)2−2(a−b)2
If x=1x−5x =\dfrac{1}{x-5}x=x−51, find :
(i) x−1xx-\dfrac{1}{x}x−x1
(ii) x+1xx+\dfrac{1}{x}x+x1
(iii) x2−1x2x^2-\dfrac{1}{x^2}x2−x21
(iv) x2+1x2x^2+\dfrac{1}{x^2}x2+x21
If x - y = 7 and x3−y3=133;x^3 - y^3 = 133;x3−y3=133; find :
(i) xy
(ii) x2+y2x^2 + y^2x2+y2
