The solution set is {2, 42} & the extraneous root is 42
What is extraneous root ?
If there are two roots in an equation, and one of them does not properly satisfy the equation, then the root is called extraneous root of that equation.
What are the roots ?
Given equation is,
[tex]2+\sqrt{2x-3}=\sqrt{x+7}[/tex]
Squaring both sides we get,
[tex]2^{2} +2(2)(\sqrt{2x-3} )+(\sqrt{2x-3}) ^{2} = (\sqrt{x+7}) ^{2}[/tex]
⇒ [tex]4+4\sqrt{2x-3} +2x-3=x+7\\[/tex]
⇒ [tex]4\sqrt{2x-3} =6-x[/tex]
Again, squaring both sides we get,
⇒ [tex](4\sqrt{2x-3}) ^{2} =(6-x)^{2}[/tex]
⇒ [tex]16(2x-3)=6^{2} -12x+x^{2}[/tex]
⇒ [tex]32x-48=36-12x+x^{2}[/tex]
⇒ [tex]x^{2} -44x+84=0[/tex]
⇒ [tex]x^{2} -42x-2x+84=0[/tex]
⇒ [tex](x-42)(x-2)=0[/tex]
So, x = 2 & 42
Here extraneous root is 42.
Learn more about extraneous root here :
https://brainly.com/question/658096
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