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Algebraic Identities

If 3x+1/23x-1...

Question

If $(3 x + \frac{1}{2}) (3 x - \frac{1}{2}) = 9 x^{2}$ -p then the value of p is
(a) 0 (b) $- \frac{1}{4}$ (c) $\frac{1}{4}$ (d) $\frac{1}{2}$

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Solution

$(3 x + \frac{1}{2}) (3 x - \frac{1}{2}) = 9 x^{2}$ -p

here $p = (\frac{1}{2})^{2} = \frac{1}{4}$

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Similar questions

(1 - p) (1 + 3 x + 9 x^{2} + 27 x^{3} + 81 x^{4} + 243 x^{5}) = 1 - p^{6}, p \neq 1

, then the value of

\frac{p}{x}

(3 x + \frac{1}{2}) (3 x - \frac{1}{2}) = 9 x^{2} - p

then the value of p is
(a) ⁰
(b)

- \frac{1}{4}

\frac{1}{4}

\frac{1}{2}

3 x - \frac{1}{3 x} = 15

, find the value.

9 x^{2} + \frac{1}{9 x^{2}}

F a c t o r i s e : 9 x^{2} - 4

3 x - \frac{1}{2 x} = 6

, then the value of

9 x^{2} + \frac{1}{4 x^{2}}

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