Solve: $25 {(2 x + y)}^{2} - 16 {(x - y)}^{2}$

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Solution

$25 (2 x + y)^{2} - 16 (x - y)^{2}$

$= 25 (4 x^{2} + y^{2} + 4 x y) - 16 (x^{2} + y^{2} - 2 x y)$

$= 100 x^{2} + 25 y^{2} + 100 x y - 16 x^{2} - 16 y^{2} + 32 x y$

$= 84 x^{2} + 9 y^{2} + 132 x y$

Hence, the value is $84 x^{2} + 9 y^{2} + 132 x y$ .

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Factorise : $25 (2 x + y)^{2} - 16 (x - y)^{2}$

Q. Factorise by using formula:

25 (2 x + y)^{2} - 16 (x - y)^{2}

Q. State whether True or False:

Factorization of

25 (2 x - y)^{2} - 16 (x - 2 y)^{2}

3 (14 x - 13 y) (2 x + y)

Factorise :

$(i) a^{2} - (b - c)^{2} (i i) 25 (2 x - y)^{2} - 16 (x - 2 y)^{2} (i i i) 16 (5 x + 4)^{2} - 9 (3 x - 2)^{2} (i v) 9 x^{2} - \frac{1}{16} (v) 25 (x - 2 y)^{2} - 4$

Q. The equation of the circle which orthogonally intersects the circles

x^{2} + y^{2} - 2 x + 3 y - 7 = 0

x^{2} + y^{2} + 5 x - 5 y + 9 = 0

and

x^{2} + y^{2} + 7 x - 9 y + 29 = 0

, is

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Solve:25(2x+y)2−16(x−y)2

25(2x+y)2−16(x−y)2=25(4x2+y2+4xy)−16(x2+y2−2xy)=100x2+25y2+100xy−16x2−16y2+32xy=84x2+9y2+132xyHence, the value is 84x2+9y2+132xy.

Solve: $25 {(2 x + y)}^{2} - 16 {(x - y)}^{2}$

$25 (2 x + y)^{2} - 16 (x - y)^{2}$

$= 25 (4 x^{2} + y^{2} + 4 x y) - 16 (x^{2} + y^{2} - 2 x y)$

$= 100 x^{2} + 25 y^{2} + 100 x y - 16 x^{2} - 16 y^{2} + 32 x y$

$= 84 x^{2} + 9 y^{2} + 132 x y$

Hence, the value is $84 x^{2} + 9 y^{2} + 132 x y$ .