Question

If x² + y² + 10 = 2√2 x + 4√2 y, then find the value of ( x + y ).

Answer 1

ANSWER

Given:

• x² + y² + 10 = 2√2 x + 4√2 y

To Find:

• Value of (x + y)

Solution:

⇒ x² + y² + 10 = 2√2 x + 4√2 y

⇒ x² + y² + 10 - 2√2 x - 4√2 y = 0

Rearranging the terms,

⇒ x² - 2√2 x + y² - 4√2 y + 10 = 0

⇒ (x)² - 2*(√2)*(x) + (y)² - 2*(2√2)*(y) + 10 = 0

Adding and subtracting both (√2)² and (2√2)²

⇒ (x)² - 2*(√2)*(x) + (y)² - 2*(2√2)*(y) + 10 + (√2)² - (√2)² + (2√2)² - (2√2)²= 0

Rearranging the terms,

⇒ [(x)² - 2*(√2)*(x) + (√2)²] + [(y)² - 2*(2√2)*(y) + (2√2)²] + [10 - (√2)² - (2√2)²]= 0

We know that,

a² - 2ab + b² = (a - b)²

So,

⇒ [x - √2]² + [y - 2√2]² + [10 - 2 - 8] = 0

⇒ [x - √2]² + [y - 2√2]² + 10 - 10 = 0

⇒ [x - √2]² + [y - 2√2]² = 0

⇒ [x - √2]² = - [y - 2√2]²

⇒ [x - √2]² = [2√2 - y]²

Square rooting both sides,

⇒ x - √2 = 2√2 - y

⇒ x + y = 2√2 + √2

⇒ x + y = 3√2

Formula Used:

• a² - 2ab + b² = (a - b)²

Learn More:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identities}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) - B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$

Answer 2

Given : $\longmapsto \bf x^2 + y^2 + 10 = 2\sqrt {2} x + 4\sqrt {2} y$

Need To Find : The value of ( x + y ) .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \quad \bf \dag \:\: x^2 + y^2 + 10 = 2\sqrt {2} x + 4\sqrt {2} y$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Solving \: the \: Given \: \::}}$

$\qquad :\implies \sf x^2 + y^2 + 10 = 2\sqrt {2} x + 4\sqrt {2} y$

Or ,

$\qquad :\implies \sf x^2 + y^2 + 10 - 2\sqrt {2} x - 4\sqrt {2} y = 0$

$\sf { By \:Rearranging \:the\:Like \:terms \:we\:get\::}$

$\qquad :\implies \sf x^2 + y^2 + 10 - 2\sqrt {2} x - 4\sqrt {2} y = 0$

$\qquad :\implies \sf x^2 -2\sqrt{2}x + y^2 - 4\sqrt{2}y + 10 = 0$

$\qquad :\implies \sf x^2 -2\times(\sqrt{2})\times(x) + y^2 - 2\times(2\sqrt{2})\times(y) + 10 = 0$

$\sf By \:Adding \:and\:Subtracting \:(\sqrt {2})^2 \:\&\: (2\sqrt {2})^2\:we\:get\::$

$\qquad :\implies \sf x^2 -2\times(\sqrt{2})\times(x) + y^2 - 2\times(2\sqrt{2})\times(y) + 10 + (\sqrt {2})^2 \:-\: (2\sqrt {2})^2 + (\sqrt {2})^2 \:-\: (2\sqrt {2})^2 = 0$

$\sf { By \:Rearranging \:the\:Like \:terms \:we\:get\::}$

$\qquad :\implies \sf x^2 -2\times(\sqrt{2})\times(x) + y^2 - 2\times(2\sqrt{2})\times(y) + 10 + (\sqrt {2})^2 \:-\: (2\sqrt {2})^2 + (\sqrt {2})^2 \:-\: (2\sqrt {2})^2 = 0$

$\qquad :\implies \sf \bigg( x^2 -2\times(\sqrt{2})\times(x) + (\sqrt {2})^2\bigg) + \bigg( y^2 - 2\times(2\sqrt{2})\times(y) + (2\sqrt {2})^2\bigg) + \bigg( 10 - (2\sqrt {2})^2 - (\sqrt {2})^2 \bigg) = 0$

$\sf { By \:Solving \:we\:get\::}$

$\qquad :\implies \sf \bigg( x^2 -2\times(\sqrt{2})\times(x) + (\sqrt {2})^2\bigg) + \bigg( y^2 - 2\times(2\sqrt{2})\times(y) + (2\sqrt {2})^2\bigg) + \bigg( 10 - 4\times2 - 2 \bigg) = 0$

$\qquad :\implies \sf \bigg( x^2 -2\times(\sqrt{2})\times(x) + (\sqrt {2})^2\bigg) + \bigg( y^2 - 2\times(2\sqrt{2})\times(y) + (2\sqrt {2})^2\bigg) + \bigg( 10 - 8 - 2 \bigg) = 0$

$\qquad :\implies \sf \bigg( x^2 -2\times(\sqrt{2})\times(x) + (\sqrt {2})^2\bigg) + \bigg( y^2 - 2\times(2\sqrt{2})\times(y) + (2\sqrt {2})^2\bigg) \cancel{+ \bigg( 10 - 10 \bigg)} = 0$

$\qquad :\implies \sf \bigg( x^2 -2\times(\sqrt{2})\times(x) + (\sqrt {2})^2\bigg) + \bigg( y^2 - 2\times(2\sqrt{2})\times(y) + (2\sqrt {2})^2\bigg) = 0$

As, We know that ,

$\bf Algebraic \: Indentity = ( a - b )^2 = a^2 - 2ab + b^2$

$\sf { By \:Using\:Indentity \:we\:get\::}$

$\qquad :\implies \sf \bigg( x - \sqrt {2}\bigg)^2 + \bigg( y - 2\sqrt {2}\bigg)^2 = 0$

Or ,

$\qquad :\implies \sf \bigg( x - \sqrt {2}\bigg)^2 = - \bigg( y - 2\sqrt {2}\bigg)^2$

$\qquad :\implies \sf \bigg( x - \sqrt {2}\bigg)^2 = \bigg( -y + 2\sqrt {2}\bigg)^2$

Or ,

$\qquad :\implies \sf \bigg( x - \sqrt {2}\bigg)^2 = \bigg( 2\sqrt {2}-y\bigg)^2$

$\sf { By \:Using\:Square \:Rooting\:both\:side\:we\:get\::}$

$\qquad :\implies \sf \sqrt {\bigg( x - \sqrt {2}\bigg)^2} = \sqrt {\bigg( 2\sqrt {2}-y\bigg)^2}$

As , We know that ,

• $\sqrt {a^{2}} = a$

$\sf { By \:Using\:this \:we\:get\::}$

$\qquad :\implies \sf x - \sqrt {2} = 2\sqrt {2}-y$

$\qquad :\implies \sf x +y = \sqrt {2} + 2\sqrt {2}$

$\qquad :\implies \bf x +y = 3\sqrt {2} \qquad \longrightarrow Required \:AnswEr \:$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  The\:Value\:of\:(x + y) \:is\:\bf{3\sqrt{2}}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

• $\boxed{\begin{array}{cc}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{array}}$

⠀⠀