raAnswer:
x = 5
Step-by-step explanation:
Original equation = [tex]\sqrt{2x+6}[/tex] - [tex]\sqrt{x+4}[/tex] = 1
Remove square roots
[tex]x^{2}[/tex] + 2x +1 = 4x +16
Subtract 16 from both sides
[tex]x^{2}[/tex]+ 2x +1 = 4x +16
-16 -16
---------------------------
[tex]x^{2}[/tex] + 2x- 15= 4x
Subtract 4x from both sides
[tex]x^{2}[/tex] + 2x - 15 - 4x= 4x - 4x
= [tex]x^{2}[/tex] - 2x - 15 = 0
THIS EQUATION HAS 2 rational roots. {x1,x2} = {5,-3}
CHECK THAT THE FIRST SOLUTION IS CORRECT (5)
Original equation, root isolated, after tidy up
[tex]\sqrt{2x+6}[/tex] = [tex]\sqrt{x+4}[/tex]
Plug in 5 for x
[tex]\sqrt{2\cdot (5)+6}[/tex] = [tex]\sqrt{(5)+4+1}[/tex]
Simplify
[tex]\sqrt16}[/tex] = 4
Solution checks !!
Solution is: x = 5
CHECK THAT THE SECOND SOLUTION IS CORRECT (-3)
Original equation, root isolated, after tidy up
[tex]\sqrt{2x+6}[/tex] = [tex]\sqrt{x+4}[/tex]
Plug in -3 for x
[tex]\sqrt{2\cdot (-3) +6}[/tex] = [tex]\sqrt{(-3)+4+1}[/tex]
Simplify
[tex]\sqrt{0}[/tex] = 2
Solution does not check
0 ≠ 2
So, the answer is x = 5
Hope this helps!
