These are 3 questions and 3 answers.
Question 1.
Answer: bot equations have the same potential solutions, but equation A may have extraneous solutions
Explanation:
1) Equation A
[tex] \sqrt{x^2+3x-6}=\sqrt{x+2}\\ \\
x^2+3x-6=x+2\\ \\
x^2+2x-8=0\\ \\
(x+4)(x-2)=0\\ \\
x=-4 ; x =2 [/tex]
Replace x = -4 in the right side of the equation:
[tex] \sqrt{x+2} =\sqrt{-4+2} =\sqrt{-2} [/tex]
Which is not defined, so this is a extraneous solution.
2) Equation B.
Since it is a cube root, it is defined for any (negative, zero, and positive) values of x.
Question 2.
Answer: cubing both sides once.
Explanation:
[tex] \sqrt[3]{5x-2} -\sqrt[3]{4x} =0\\ \\
\sqrt[3]{5x-2} =\sqrt[3]{4x}\\ \\
(\sqrt[3]{5x-2} )^3=(\sqrt[3]{4x})^3\\ \\
5x-2=4x\\ \\
5x-4x=2\\ \\
x=2 [/tex]
3) Question 3.
Answer: c - 2 = 25 + c + 10√c
Explanation:
[tex] \sqrt{c-2} -\sqrt{c} =5\\ \\ \\
\sqrt{c-2} =\sqrt{c} +5\\ \\
(\sqrt{c-2})^{2} =(\sqrt{c}+5)^2\\ \\
c-2=c+10\sqrt{c} +25\\ \\
c-2=25+c+10\sqrt{c} [/tex]
