Question

The following radical equation has a solution set {x = ?} and an extraneous root x = ?.

Answer 1

Answer:

x=2; the extraneous root x=42.

All the details are in the attached picture, the answer is marked with red colour.

Answer 2

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,

$2+\sqrt{2x-3}=\sqrt{x+7}$

Squaring both sides we get,

$2^{2} +2(2)(\sqrt{2x-3} )+(\sqrt{2x-3}) ^{2} = (\sqrt{x+7}) ^{2}$

⇒ $4+4\sqrt{2x-3} +2x-3=x+7$

⇒ $4\sqrt{2x-3} =6-x$

Again, squaring both sides we get,

⇒ $(4\sqrt{2x-3}) ^{2} =(6-x)^{2}$

⇒ $16(2x-3)=6^{2} -12x+x^{2}$

⇒ $32x-48=36-12x+x^{2}$

⇒ $x^{2} -44x+84=0$

⇒ $x^{2} -42x-2x+84=0$

⇒ $(x-42)(x-2)=0$

So, x = 2 & 42

Here extraneous root is 42.

Learn more about extraneous root here :

https://brainly.com/question/658096

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