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Algebraic Identities

If x +2p+3q...

Question

If $x + 2 p + 3 q = 14$ and $2 x p + 6 p q + 3 q x = 4$ , find the value of $x^{2} + 4 p^{2} + 9 q^{2} .$

A

199

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B

132

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C

188

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D

178

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Solution

The correct option is C $188$
Given,
$x + 2 p + 3 q = 14$ and $2 x p + 6 p q + 3 q x = 4$
We know,
$(a + b + c)^{2} = (a^{2} + b^{2} + c^{2}) + 2 (a b + b c + c a)$
Here,
$a = x, b = 2 p, c = 3 q$
Thus,
$(x + 2 p + 3 q)^{2} = x^{2} + 4 p^{2} + 9 q^{2} + 2 (2 x p + 6 p q + 3 q x)$
$=> 14^{2} = x^{2} + 4 p^{2} + 9 q^{2} + 2 (4)$
$=> 196 - 8 = x^{2} + 4 p^{2} + 9 q^{2}$
$=> x^{2} + 4 p^{2} + 9 q^{2} = 188$

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Similar questions

x = \frac{\sqrt{2 p + 3 q} + \sqrt{2 p - 3 q}}{\sqrt{2 p + 3 q} - \sqrt{2 p - 3 q}}

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\frac{x + 3 p}{x - 3 p} + \frac{x + 3 q}{x - 3 q}

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then find value of

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