Question 79

If $$16qxy = (2x + y)^{2} - (2x - y)^{2}$$, then what is the value of q?

Solution

Expression : $$16qxy = (2x + y)^{2} - (2x - y)^{2}$$

R.H.S. = $$(2x + y)^2 - (2x - y)^2$$

= $$(4x^2 + y^2 + 4xy) - (4x^2 + y^2 - 4xy)$$

= $$(4x^2 - 4x^2) + (y^2 - y^22) + (4xy + 4xy)$$

= $$8xy$$

L.H.S. = $$16qxy$$

=> $$16q = 8$$

=> $$q = \frac{8}{16} = 0.5$$

=> Ans - (A)

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