Question

Solve root(x+5) +root(x+12)=root2x+41..

Answer 1

SOLUTION

Here given Equation is

$\sqrt{(x + 5)} + \sqrt{(x + 12)} = \sqrt{2x + 41}$

Squaring both the sides, we get

$= > (x + 5) + (x + 12) + 2 \sqrt{(x + 5)(x + 12)} =2x + 41 \\ = > 2 \sqrt{(x + 5)(x + 12)} = 24 \\ = > \sqrt{(x + 5)(x + 12)} = 12$

Again squaring both the sides, we get

$= > (x + 5)(x + 12) = 144 \\ = > {x}^{2} + 17x + 60 = 144 \\ = > {x}^{2} + 17x - 84 = 0 \\ = > {x}^{2} + 21x - 4x - 84 = 0 \\ = > x(x + 21) - 4(x + 21) = 0 \\ = > (x + 21)(x - 4) = 0 \\ so \\ = >x = - 21 \: \: \: and \: \: \: x = 4$

Solving for x = 4 is a root, because it satisfy the given Equation. Also x= -21 is an extraneous root because it does not satisfy the given Equation.

hope it helps ✔️

Answer 2

root(x+5) +root(x+12)=root2x+41..

squaring on the both sides

(x+5) + 2root(x+5)root(x+12) + x+12 = 2x +41

2x-2x +17 -41 +2root (x+5)root(x+12) =0

2root (x+5)root(x+12) = 24

squaring on the both sides

4(x+5)(x+12) = (24)^2

4(x^2 + 12x +5x+60) = 576

4x^2 + 68x + 240 = 576

4x^2 + 68x - 336 = 0

ie 4( x^2 + 17x - 84) = 0

so, x^2 + 17x - 84 = 0

Now

by using quadratic

x = -b +_ √(b^2 - 4ac) \ 2

where

a = 1 , b= 17 , c = -84

x = {-(17) +_ √((17) ^2- 4x1x( -84) )}\ 2

x = -17 +_ √(289 + 336) \ 2

x ={ - 17 +_ root (625) }

x = -17 - 25\ 2 [;, √625 = 25]

taking subtraction ( - )

x = -17 - 25\ 2 [;, √625 = 25]

x = -41 \2

taking (+) we get.

x = - 17 + 25\2

= 8\2

=4

i. e not possible to write negative value.(- 41\2)

so, the possible value of x = 4