Solving $\sqrt{4x+13}=x+2$ produces the candidates x = 3 and x = -3. Which value is extraneous?

After solving: the cube root of (x + 3) equals 2, why is no check for extraneous solutions needed?

Squaring both sides of $\sqrt{x+20}=x$ gives the candidates x = 5 and x = -4. Substituting x = -4 makes the left side 4 and the right side -4. What does this show?

Solving $\sqrt{6x+25}=x+3$ produces the candidates x = 4 and x = -4. Which value is extraneous?

After solving: the fifth root of (x + 1) equals 2, why is no check for extraneous solutions needed?

After solving: the cube root of (3x + 1) equals 2, why is no check for extraneous solutions needed?

After solving: the fifth root of (x + 5) equals 2, why is no check for extraneous solutions needed?

After solving: the cube root of (2x + 4) equals 2, why is no check for extraneous solutions needed?

Solving $\sqrt{5x+26}=x+4$ produces the candidates x = 2 and x = -5. Which value is extraneous?

Solving $\sqrt{2x+37}=x+1$ produces the candidates x = 6 and x = -6. Which value is extraneous?

Squaring both sides of $\sqrt{x+6}=x$ gives the candidates x = 3 and x = -2. Substituting x = -2 makes the left side 2 and the right side -2. What does this show?

Squaring both sides of $\sqrt{x+42}=x$ gives the candidates x = 7 and x = -6. Substituting x = -6 makes the left side 6 and the right side -6. What does this show?

After solving: the cube root of (3x + 2) equals 2, why is no check for extraneous solutions needed?

Squaring both sides of $\sqrt{x+12}=x$ gives the candidates x = 4 and x = -3. Substituting x = -3 makes the left side 3 and the right side -3. What does this show?

Squaring both sides of $\sqrt{x+72}=x$ gives the candidates x = 9 and x = -8. Substituting x = -8 makes the left side 8 and the right side -8. What does this show?

Solving $\sqrt{5x+34}=x+4$ produces the candidates x = 3 and x = -6. Which value is extraneous?

Solving $\sqrt{2x+12}=x+2$ produces the candidates x = 2 and x = -4. Which value is extraneous?

Solving $\sqrt{6x+19}=x+2$ produces the candidates x = 5 and x = -3. Which value is extraneous?

Squaring both sides of $\sqrt{x+2}=x$ gives the candidates x = 2 and x = -1. Substituting x = -1 makes the left side 1 and the right side -1. What does this show?

Squaring both sides of $\sqrt{x+56}=x$ gives the candidates x = 8 and x = -7. Substituting x = -7 makes the left side 7 and the right side -7. What does this show?