Radical Equation Solver: √(ax + b) = c
Enter the coefficient 'a' for 'x' inside the radical.
Enter the constant 'b' inside the radical.
Enter the constant 'c' on the right side of the equation.
Result will appear here.
function calculateRadicalEquation() {
var coeffA = parseFloat(document.getElementById('coeffA').value);
var constB = parseFloat(document.getElementById('constB').value);
var valueC = parseFloat(document.getElementById('valueC').value);
var resultDiv = document.getElementById('radicalResult');
if (isNaN(coeffA) || isNaN(constB) || isNaN(valueC)) {
resultDiv.innerHTML = 'Please enter valid numbers for all fields.';
return;
}
if (coeffA === 0) {
resultDiv.innerHTML = 'Error: Coefficient \'a\' cannot be zero for this type of solution.';
return;
}
if (valueC < 0) {
resultDiv.innerHTML = 'No real solution: The right side of a square root equation (c) cannot be negative.';
return;
}
// Square both sides: ax + b = c^2
// Isolate ax: ax = c^2 – b
// Solve for x: x = (c^2 – b) / a
var cSquared = valueC * valueC;
var numerator = cSquared – constB;
var x = numerator / coeffA;
// Verification step: Check if the expression under the radical is non-negative for the calculated x
var checkUnderRadical = coeffA * x + constB;
if (checkUnderRadical 1e-9) { // Allow for small floating point inaccuracies
resultDiv.innerHTML = 'No real solution: The value under the radical (ax + b) would be negative for x = ' + x.toFixed(4) + '. This is an extraneous solution.';
return;
}
// Also verify that sqrt(ax+b) actually equals c (to catch extraneous solutions if c was positive but the original radical was negative)
// For example, sqrt(x) = -2 has no solution, but squaring gives x=4. sqrt(4) is 2, not -2.
// Our check for valueC 1e-9) {
resultDiv.innerHTML = 'No real solution: The calculated x does not satisfy the original equation. This is an extraneous solution.';
return;
}
resultDiv.innerHTML = 'Solution for x: x = ' + x.toFixed(4) + '';
}
Understanding Radical Equations
A radical equation is an equation in which the variable appears under a radical sign, most commonly a square root. These equations are fundamental in algebra and are used in various fields, including physics, engineering, and finance, to model situations where quantities are related by roots.
The Equation This Calculator Solves
This calculator is designed to solve radical equations of the form:
√(ax + b) = c
Where:
- a is the coefficient of the variable 'x' inside the square root.
- b is the constant term inside the square root.
- c is the constant term on the right side of the equation.
How to Solve √(ax + b) = c
The general strategy to solve a radical equation is to isolate the radical term and then eliminate the radical by raising both sides of the equation to the power corresponding to the index of the radical (e.g., square both sides for a square root, cube both sides for a cube root).
- Isolate the Radical: In our specific form, √(ax + b) = c, the radical is already isolated on one side.
- Square Both Sides: To eliminate the square root, we square both sides of the equation:
(√(ax + b))2 = c2
ax + b = c2
- Solve for x: Now, we have a linear equation.
ax = c2 – b
x = (c2 – b) / a
- Check for Extraneous Solutions: This is a crucial step for radical equations. Squaring both sides can sometimes introduce "extraneous solutions" that do not satisfy the original equation. You must always substitute your calculated 'x' back into the original equation to verify it.
- The value under the radical (ax + b) must be non-negative.
- The right side of the equation (c) must be non-negative, as a square root of a real number cannot result in a negative value.
Examples
Example 1: Simple Case
Solve: √(2x + 1) = 3
- Here, a = 2, b = 1, c = 3.
- Square both sides: 2x + 1 = 32
- 2x + 1 = 9
- 2x = 8
- x = 4
- Check: √(2*4 + 1) = √(8 + 1) = √9 = 3. This matches the original equation. So, x = 4 is the correct solution.
Example 2: No Real Solution (Negative 'c')
Solve: √(x + 5) = -2
- Here, a = 1, b = 5, c = -2.
- Since the right side (c) is negative, and a square root of a real number cannot be negative, there is no real solution.
- If we were to proceed by squaring: x + 5 = (-2)2 → x + 5 = 4 → x = -1.
- Check: √(-1 + 5) = √4 = 2. This does not equal -2. Thus, x = -1 is an extraneous solution.
Example 3: No Real Solution (Extraneous)
Solve: √(3x – 2) = 1
- Here, a = 3, b = -2, c = 1.
- Square both sides: 3x – 2 = 12
- 3x – 2 = 1
- 3x = 3
- x = 1
- Check: √(3*1 – 2) = √(3 – 2) = √1 = 1. This matches the original equation. So, x = 1 is the correct solution.