Show that:
$(9 p - 5 q)^{2} + 180 p q = (9 p + 5 q)^{2}$

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Solution

L.H.S = $(9 p - 5 q)^{2} + 180 p q$
$= (9 p)^{2} - 2 \times 9 p \times 5 q + (5 q)^{2} + 180 p q$
$[U s i n g i d e n t i t y (a - b)^{2} = a^{2} - 2 a b + b^{2}]$
$= 81 p^{2} - 90 p q + 25 q^{2} + 180 p q$
$= 81 p^{2} + 90 p q + 25 q^{2}$
$= (9 p)^{2} + 2 \times 9 p \times 5 q + (5 q)^{2}$
$= (9 p + 5 q)^{2}$ $[∵ (a + b)^{2} = a^{2} + 2 a b + b^{2}]$

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(9 p - 5 q)^{2} + 180 p q = (9 p + 5 q)^{2}

Show that:
$(9 p - 5 q)^{2} + 180 p q = (9 p + 5 q)^{2}$

Question 5(ii)
Show that:
$(9 p - 5 q)^{2} + 180 p q = (9 p + 5 q)^{2}$

(9 p - 5 q)^{2} + 180 p q

The value of $(9 p - 5 q)^{2} + 180 p q =$

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