Question

set of values of x,
$\sqrt{(x + 8) } + \sqrt{(2x + 2)} = 1$
is

Answer 1

Solution :

√x+8 + √(2x+2) = 1

=> √(2x+2) = 1 - √(x+8)

Do the square on both sides,we get

=> 2x+2 = 1 + x + 8 - 2√(x+8)

=> 2x + 2 -1 - x - 8 = -2√(x+8)

=> x - 7 = -2√(x+8)

Do the square both sides , we get

=> ( x - 7 )² = [ -2(√x+8 ) ]²

=> x² + 49 - 14x = 4( x + 8 )

=> x² - 14x + 49 - 4x - 32 = 0

=> x² - 18x + 17 = 0

=> x² - x - 17x + 17 = 0

=> x( x - 1 ) - 17( x - 1 ) = 0

=> ( x - 1 )( x - 17 ) = 0

Therefore ,

x - 1 = 0 or x - 17 = 0

x = 1 or x = 17

•••••

Answer 2

Answer : 17 and 1

Solution :
_________

Given that :

$\sqrt{x + 8} + \sqrt{2x + 2} = 1$

On solving the equation :

$\sqrt{(x + 8)} = 1 - \sqrt{(2x + 2)} \\ \\ On \: squaring \: both \: the \: side: \: \\ \\ = > x + 8 = 1 - 2 \sqrt{2x + 2} + 2x + 2 \\ \\ = > x - 5 = 2 \sqrt{2x + 2} \\ \\ Again \: squairing \: both \: the \: side: \\ \\ = > {x}^{2} - 10x + 25 = 4(2x + 2) \\ \\ = > {x}^{2} - 10x - 8x + 25 - 8 = 0 \\ \\ = > {x}^{2} - 18x + 17 = 0 \\ \\ = > {x}^{2} - 17x - x + 17 = 0 \\ \\ = > x(x - 17) - 1(x - 17) = 0 \\ \\ = > (x - 17)(x - 1) = 0$

So, required value of x will be 1 and 17