If b-1b-4- then b2-1b2-

The correct option is D 14 Here, b + 1b = 4 Squaring on both sides #160; ( b + 1b)^2 = 4^2 b^2 + 1b^2 + 2 × b ×1b = 16 b^2 + 1b^2 + 2 ...

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Standard VII

Mathematics

(a + b)^2 Expansion and Visualisation

If b+1b=4, ...

Question

If $b + \frac{1}{b} = 4$ , then $b^{2} + \frac{1}{b^{2}}$ =__________.

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Solution

The correct option is D $14$
Here, $b + \frac{1}{b} = 4$

Squaring on both sides
${(b + \frac{1}{b})}^{2} = 4^{2}$

$b^{2} + \frac{1}{b^{2}} + 2 \times b \times \frac{1}{b} = 16$

$b^{2} + \frac{1}{b^{2}} + 2 = 16$

$b^{2} + \frac{1}{b^{2}} = 14$

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If b+1b=4, then b2+1b2=__________.

The correct option is D 14Here, b+1b=4Squaring on both sides (b+1b)2=42b2+1b2+2×b×1b=16b2+1b2+2=16b2+1b2=14

If $b + \frac{1}{b} = 4$ , then $b^{2} + \frac{1}{b^{2}}$ =__________.

The correct option is D $14$
Here, $b + \frac{1}{b} = 4$

Squaring on both sides
${(b + \frac{1}{b})}^{2} = 4^{2}$

$b^{2} + \frac{1}{b^{2}} + 2 \times b \times \frac{1}{b} = 16$

$b^{2} + \frac{1}{b^{2}} + 2 = 16$

$b^{2} + \frac{1}{b^{2}} = 14$