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

$(9 p + 7 q)^{2}$

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

$(9 p - 5 q)^{2}$

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

$(8 p + 5 q)^{2}$

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

Open in App

Solution

The correct option is C
$(9 p + 5 q)^{2}$

$(9 p - 5 q)^{2} + 180 p q = 81 p^{2} - 90 p q + 25 q^{2} + 180 p q = 81 p^{2} + 90 p q + 25 q^{2} = (9 p + 5 q)^{2}$

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

(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