Algebra Calculator

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Algebra Calculator
Linear Equation (ax + b = c)Quadratic Equation (ax² + bx + c = 0)
Solution:
Enter values and click Solve to see results.
function updateLabels(){var type=document.getElementById('calc_type').value;if(type=='linear'){document.getElementById('label_a').innerHTML='Coefficient (a):';document.getElementById('label_b').innerHTML='Constant (b):';document.getElementById('label_c').innerHTML='Total (c):';}else{document.getElementById('label_a').innerHTML='Coefficient (a):';document.getElementById('label_b').innerHTML='Coefficient (b):';document.getElementById('label_c').innerHTML='Constant (c):';}}function resetAnswer(){document.getElementById('answer').innerHTML='Enter values and click Solve to see results.';}function calculateAlgebra(){var a=parseFloat(document.getElementById('val_a').value);var b=parseFloat(document.getElementById('val_b').value);var c=parseFloat(document.getElementById('val_c').value);var type=document.getElementById('calc_type').value;var showSteps=document.getElementById('show_steps').checked;var ansDiv=document.getElementById('answer');if(isNaN(a)||isNaN(b)||isNaN(c)){ansDiv.innerHTML='Please enter valid numeric values.';return;}if(type=='linear'){if(a==0){ansDiv.innerHTML='Error: Coefficient (a) cannot be zero in a linear equation.';return;}var x=(c-b)/a;var output='x = '+x.toFixed(4)+'';if(showSteps){output+='
Step 1: Subtract '+b+' from both sides: '+a+'x = '+(c-b)+'
Step 2: Divide by '+a+': x = '+(c-b)+'/'+a+'
Final Answer: x = '+x.toFixed(4)+'
';}ansDiv.innerHTML=output;}else{if(a==0){ansDiv.innerHTML='Error: "a" cannot be zero for a quadratic equation. Use Linear mode.';return;}var disc=(b*b)-(4*a*c);var output=";if(disc>0){var x1=(-b+Math.sqrt(disc))/(2*a);var x2=(-b-Math.sqrt(disc))/(2*a);output='x₁ = '+x1.toFixed(4)+'
x₂ = '+x2.toFixed(4)+'';if(showSteps){output+='
Step 1: Discriminant (D) = b² – 4ac = '+disc+'
Step 2: Since D > 0, there are two real roots.
Step 3: Apply formula [-b ± √D] / 2a
';}}else if(disc==0){var x=-b/(2*a);output='x = '+x.toFixed(4)+' (Repeated Root)';}else{var realPart=(-b/(2*a)).toFixed(4);var imagPart=(Math.sqrt(-disc)/(2*a)).toFixed(4);output='x₁ = '+realPart+' + '+imagPart+'i
x₂ = '+realPart+' – '+imagPart+'i';}ansDiv.innerHTML=output;}}

Algebra Calculator Use

The algebra calculator is a versatile tool designed to solve fundamental algebraic equations quickly and accurately. Whether you are a student working on homework or a professional needing to isolate a variable, this calculator handles both linear and quadratic equations. By simply inputting the known coefficients, you can find the value of the unknown variable (x) without performing manual arithmetic.

To use this tool, select the type of equation you wish to solve from the dropdown menu. The interface will automatically update the labels to guide your data entry.

Coefficient (a)
The number multiplying the variable (x or x²). In algebra, this is the leading coefficient.
Coefficient/Constant (b)
In a linear equation, this is the constant added to the variable. In a quadratic equation, this is the coefficient of the x term.
Constant (c)
The independent number in the equation. For quadratic equations, this is the term without a variable.

How It Works

Algebra relies on balancing equations. The algebra calculator uses two primary mathematical frameworks depending on your selection:

Linear Equations (ax + b = c)

For linear equations, the goal is to isolate x. The calculator performs the following operations:

x = (c – b) / a

Quadratic Equations (ax² + bx + c = 0)

Quadratic equations are solved using the Standard Quadratic Formula. This accounts for scenarios with two real roots, one repeated root, or complex (imaginary) roots:

x = [-b ± √(b² – 4ac)] / 2a

Algebra Calculation Examples

Example 1: Linear Solution
Solve for x: 5x + 10 = 30.

  1. Set a = 5, b = 10, c = 30
  2. Subtract 10 from both sides: 5x = 20
  3. Divide by 5: x = 4
  4. Result: x = 4.00

Example 2: Quadratic Solution
Solve for x: x² – 5x + 6 = 0.

  1. Set a = 1, b = -5, c = 6
  2. Calculate Discriminant: (-5)² – 4(1)(6) = 25 – 24 = 1
  3. Apply Formula: [5 ± √1] / 2
  4. x₁ = (5+1)/2 = 3; x₂ = (5-1)/2 = 2
  5. Result: x₁ = 3.00, x₂ = 2.00

Common Questions

Can this algebra calculator solve for y?

While the labels use 'x', algebra is variable-neutral. You can treat 'x' as 'y', 'z', or any other variable you are solving for, as long as the equation follows the linear or quadratic structure.

What happens if 'a' is zero?

In a linear equation, if 'a' is zero, there is no variable to solve for, resulting in an undefined operation. In a quadratic equation, if 'a' is zero, the equation becomes linear (bx + c = 0), and you should use the Linear Equation setting.

Does it handle negative numbers?

Yes. You can enter negative values for any coefficient. For example, to solve x² – 4 = 0, you would enter a=1, b=0, and c=-4.

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