$(9 p - 5 q)^{2} + 180 p q$ is equivalent to _______.

(5 p + 9 q)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(5 p - 9 q)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(9 p + 5 q)^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(9 p - 5 q)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(9 p + 5 q)^{2}$
We know that $a^{2} + b^{2} + 2 a b$
$(a - b)^{2} = a^{2} + b^{2} - 2 a b$
$\Rightarrow (a - b)^{2} + 4 a b = a^{2} + b^{2} - 2 a b + 4 a b = (a + b)^{2}$ $(∵ a = 9 p$ and $b = 5 q)$
$\Rightarrow 4 a b = 4 (9 p) (5 q) = 180 p q$
Therefore, $(9 p - 5 q)^{2} + 180 p q = (9 p + 5 q)^{2}$

Option (C) is correct.

Suggest Corrections

Similar questions

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

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

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

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

(3 x + 7)^{2} - 84 x = (3 x - 7)^{2}

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

{(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}

(4 p q + 3 q)^{2} - (4 p q - 3 q)^{2} = 48 p q^{2}

(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0

Join BYJU'S Learning Program