If $(x - 2 y + 3)^{2} + (x + 2 y - 3)^{2} = 0$ . Find $x - 6 y$

0

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x^{2} + 4 y^{2}

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- 9

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

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Solution

The correct option is C $- 9$
$(x - 2 y + 3)^{2} + (x + 2 y - 3)^{2} = 0$

if $a^{2} + b^{2} = 0$

then $a = 0, b = 0$

$x - 2 y + 3 = 0$ (1)

$x + 2 y - 3 = 0$ (2)

$(1) + (2)$

$2 x = 0$

$x = 0$ putting in (1)

$3 - 2 y = 0$

$y = \frac{3}{2}$

$x - 6 y = 0 - 6 \times \frac{3}{2} = - 9$

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Similar questions

(i) $x^{2} + y^{2} + z^{2} - x y + x + y z by x + y - z$

(ii) $x^{2} + 4 y^{2} + z^{2} + 2 x y + x z - 2 y z by x - 2 y - z$

(iii) $x^{2} + 4 y^{2} + 2 x y - 3 x + 6 y + 9 by x - 2 y + 3$

(iv) $9 x^{2} + 25 y^{2} + 15 x y + 2 x - 20 y + 16 by 3 x - 5 y + 4$

y^{2} = 4 a x

x^{2} = 32 a y

(x - 2 y)^{3} + (2 y - 32)^{3} + (32 - x)^{3} - 3 ((x - 2 y) (2 y - 32) (32 - x))

(x - 1)^{2} + (y - 3)^{2} = r^{2}

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

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