Solve Equation with a Radical | Step-by-Step Calculator

An equation of the form \( \sqrt{ax + b} + c = d \) can be solved by isolating the radical, squaring both sides, and solving the resulting linear equation.

\( \sqrt{ax + b} + c = d \) → \( \sqrt{ax + b} = d - c \) → \( ax + b = (d - c)^2 \)

⚠️ Always verify the solution because squaring can introduce extraneous solutions.

📌 The graph below shows the left side \( \sqrt{ax+b}+c \) (green) and right side \( d \) (blue). The solution is the x-coordinate of their intersection (red point).

✧ Radical Equation Solver ✧

Equation form: √(ax + b) + c = d

Enter coefficients: \( \sqrt{ax + b} + c = d \)

\( a \) (coefficient of x)

\( b \) (constant under radical)

\( c \) (constant added)

\( d \) (right side)

\( \sqrt{ax + b} + c = d \) → \( \sqrt{2x + 3} + 1 = 5 \)

⚠️ a cannot be zero. The expression under the radical must be defined (ax + b ≥ 0).

📐 Solution (Exact Value)

\( x = 9 \)

📖 Step-by-Step Solution

STEP 1: Isolate the radical term

\( \sqrt{2x + 3} = 4 \)

STEP 2: Square both sides of the equation

\( 2x + 3 = 16 \)

STEP 3: Solve the linear equation

\( x = \frac{13}{2} = 6.5 \)

STEP 4: Verify the solution (check for extraneous solutions)

\( \sqrt{2(6.5)+3} + 1 = \sqrt{16} + 1 = 4 + 1 = 5 \) ✓ Solution is valid.

💡 Important Note

Squaring both sides can introduce extraneous solutions. Always verify your solution in the original equation. If \( d - c < 0 \), the equation has no solution because a square root cannot be negative.

📊 Interactive Graph

y = √(ax+b) + c y = d (horizontal line) Intersection (solution)