If $2 x^{2} + 2 y^{2} = 640$ and $x + y = 18$ , what is the value of $x y$ ?

2

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

4

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

6

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

8

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

Open in App

Solution

The correct option is A $2$
$2 x^{2} + 2 y^{2} = 640$
$x^{2} + y^{2} = 320$ ...(i)
It is given that $(x + y) = 18$
Hence, $2 x y = (x + y)^{2} - (x^{2} + y^{2}) = 18^{2} - 320 = 324 - 320 = 4$
Hence, $2 x y = 4$ or $x y = 2$ .

Suggest Corrections

Similar questions

x - 2 y

2 x - y

2 x^{2} - x y + 2 y^{2}

x^{2} + x y + x = 12

y^{2} + x y + y = 18

x + y

x, y > 0

2 x^{2} + \frac{2}{x} - 2 x + 2 y^{2} + \frac{2}{y} - 2 y + 2

\frac{4^{x}}{2^{(x + y)}} = 8

\frac{9^{y}}{3^{5 y}} = \frac{1}{27}

(6 x y + 3 y)^{2} - (6 x y - 3 y)^{2}

Join BYJU'S Learning Program