If a + b = 4 and ab = 3; find 1b2+1a2\dfrac{1}{b^2} + \dfrac{1}{a^2}b21+a21
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Given, a + b = 4 and ab = 3.
We need to find the value of:
1b2+1a2=a2+b2a2b2=a2+b2+2ab−2aba2b2=(a2+b2+2ab)−2aba2b2=(a+b)2−2aba2b2=(a+b)2−2ab(ab)2\dfrac{1}{b^2} + \dfrac{1}{a^2}\\[1em] = \dfrac{a^2 + b^2}{a^2b^2}\\[1em] = \dfrac{a^2 + b^2 + 2ab - 2ab}{a^2b^2}\\[1em] = \dfrac{(a^2 + b^2 + 2ab) - 2ab}{a^2b^2}\\[1em] = \dfrac{(a + b)^2 - 2ab}{a^2b^2}\\[1em] = \dfrac{(a + b)^2 - 2ab}{(ab)^2}b21+a21=a2b2a2+b2=a2b2a2+b2+2ab−2ab=a2b2(a2+b2+2ab)−2ab=a2b2(a+b)2−2ab=(ab)2(a+b)2−2ab
Putting the value of (a + b) and ab,
=42−2×332=16−69=109=119= \dfrac{4^2 - 2 \times 3}{3^2}\\[1em] = \dfrac{16 - 6}{9}\\[1em] = \dfrac{10}{9}\\[1em] = 1\dfrac{1}{9}=3242−2×3=916−6=910=191
Hence, 1b2+1a2=119\dfrac{1}{b^2} + \dfrac{1}{a^2} = 1\dfrac{1}{9}b21+a21=191.
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