Question

Algebra 2 help. Solving equations containing two radicals

Answer 1

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

$\sqrt{x^2+3x-6}=\sqrt{x+2}\\ \\ x^2+3x-6=x+2\\ \\ x^2+2x-8=0\\ \\ (x+4)(x-2)=0\\ \\ x=-4 ; x =2$

Replace x = -4 in the right side of the equation:

$\sqrt{x+2} =\sqrt{-4+2} =\sqrt{-2}$

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:

$\sqrt[3]{5x-2} -\sqrt[3]{4x} =0\\ \\ \sqrt[3]{5x-2} =\sqrt[3]{4x}\\ \\ (\sqrt[3]{5x-2} )^3=(\sqrt[3]{4x})^3\\ \\ 5x-2=4x\\ \\ 5x-4x=2\\ \\ x=2$

3) Question 3.

Answer: c - 2 = 25 + c + 10√c

Explanation:

$\sqrt{c-2} -\sqrt{c} =5\\ \\ \\ \sqrt{c-2} =\sqrt{c} +5\\ \\ (\sqrt{c-2})^{2} =(\sqrt{c}+5)^2\\ \\ c-2=c+10\sqrt{c} +25\\ \\ c-2=25+c+10\sqrt{c}$