HSA.REI.A.2 Tutor Intake — Rational and Radical Equations

A student is solving the rational equation $\frac{x}{x-3} = \frac{4}{x-3}$.

Before doing any algebra, what value of $x$ must be excluded from the
solution set?

A student solves a radical equation and obtains $x = 4$ and $x = -1$.
After checking in the original equation:

• $x = 4$: left side $= 3$, right side $= 3$ ✓
• $x = -1$: left side $= 2$, right side $= -1$ ✗

What is the correct solution set?

Why can squaring both sides of a radical equation introduce an extraneous
solution?

Solve the radical equation $\sqrt{x + 5} = 4$.

After solving and checking your answer in the original equation, enter
the value of $x$.

A student tries to solve $\sqrt{x} + 3 = 7$ by squaring both sides immediately,
writing $x + 9 = 49$.

The student's error is that they squared before   ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲   . The correct
first step is to subtract 3 from both sides to get $\sqrt{x} = \underline{\hspace{5em}}$.

Solve the rational equation $\frac{3}{x} = \frac{1}{2}$.

Enter the value of $x$.