Question

Given that


3p+q = 7,


find the value of
18p^{2}+ 12pq +2q^{2}

Answer 1

3p+q = 7

(3p+q)² = 7²

9p²+6pq+q² = 49

18p²+12pq+2q² = 2(9p²+6pq+q²)

                          = 2*49 = 98

I used the identity (a+b)² = a²+2ab+b²

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Answer 2

Answer:

HEY MATE,

THE ANSWER TO YOUR QUESTION IS :

$49 \sqrt{2}$ and 98

Step-by-step explanation:

$18 {p}^{2} + 12pq + 2 {q}^{2} \\ = {(3 \sqrt{2}p) }^{2} + 2(3 \sqrt{2}p )( \sqrt{2q}) + {( \sqrt{2}q) }^{2} \\ = {(3 \sqrt{2}p + \sqrt{2}q)}^{2} \\ = (3 \sqrt{2} p + \sqrt{2} q)(3 \sqrt{2} p + \sqrt{2} q) \\ = \sqrt{2} (3p + q)(3p + q)$

BECAUSE 3p+q =7

$= \sqrt{2} (7)(7) \\ = \sqrt{2} (49) \\ = 49 \sqrt{2} \: \: \: \: : answer$

THIS WAS FIRST METHOD,,

SECOND METHOD IS THIS,,

3p+q=7

(3p+q)^2 =(7)^2

(3p) ^2 + 2(3p)(q) + (q) ^2 =7×7

9p^2 + 6pq + q^2 =49

18p^2 + 12pq +2q^2

= 2(9p^2 +6pq + q^2)

= 2(49)

= 98 answer

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