Answer: The equation has a solution .
Step-by-step explanation:
Since we have given that
[tex]\sqrt{x+1}=7-2\sqrt{x}[/tex]
Squaring on both the sides, we get that,
[tex]x+1=(7-2\sqrt{x})^2\\\\x+1=49+4x-28\sqrt{x}\\\\3x+48-28\sqrt{x}=0[/tex]
Let [tex]\sqrt{x}=y\implies x=y^2[/tex]
So, our equation becomes,
[tex]3y^2-28y+48=0\\\\y=\dfrac{14}{3}\pm \dfrac{2\sqrt{13}}{3}\\\\\sqrt{x}=\dfrac{14}{3}\pm \dfrac{2\sqrt{13}}{3}\\\\x=(\dfrac{14}{3}\pm \dfrac{2\sqrt{13}}{3})^2[/tex]
So, [tex]x=\dfrac{248}{9}\pm \dfrac{56\sqrt{13}}{9}[/tex]
Hence, the equation has a solution .
