Answer:
Given equation:
[tex]\sqrt{30-2x} =x-3[/tex]
Square both sides:
[tex]\implies (\sqrt{30-2x})^2 =(x-3)^2[/tex]
[tex]\implies 30-2x =(x-3)^2[/tex]
Expand the right side:
[tex]\implies 30-2x =x^2-6x+9[/tex]
Add [tex]2x[/tex] to both sides:
[tex]\implies 30 =x^2-4x+9[/tex]
Subtract 30 from both sides:
[tex]\implies x^2-4x-21=0[/tex]
Factor:
[tex]\implies (x-7)(x+3)=0[/tex]
Therefore,
[tex]x-7=0 \implies x =7[/tex]
[tex]x+3=0 \implies x=-3[/tex]
Substitute both values of [tex]x[/tex] into the original equation to check:
[tex]x=7 :[/tex]
[tex]\implies\sqrt{30-2(7)} =7-3[/tex]
[tex]\implies 4=4[/tex]
correct!
[tex]x=-3[/tex]
[tex]\implies\sqrt{30-2(-3)} =-3-3[/tex]
[tex]\implies 6 \neq =-6[/tex]
incorrect!
Therefore, [tex]x=7[/tex] is the only correct solution, and [tex]x=-3[/tex] is the extraneous solution
**Extraneous solution: a root of a transformed equation that is not a root of the original equation**
