Completing the Square Calculator

How to Complete the Square Using This Calculator

• Fill out the input field with the quadratic expression you want to solve.
• Click the solve button to initiate the calculation process.
• The calculator will perform the completing the square method and provide the result.
• The output will display the variable value for the specified term in the quadratic equation.

By following these simple steps, you can easily utilize the power of this calculator to complete the square and find the variable value for your quadratic equation.

What is Completing the Square?

The completing the square method is used in mathematics to solve quadratic equations. It transforms the given equation into a form where the left side becomes a perfect square binomial. This method applies when the quadratic equation is \( ax^2+ bx+ c=0 \). By completing the square, we can find the solutions to quadratic equations that cannot be easily solved using quadratic formulas.

This approach is handy in cases where the quadratic equation needs to factor more easily or when complex numbers are involved. By rearranging the equation and completing the square, we can derive a simpler form to determine the variable values and find the solutions more efficiently.

Observe that complete squares such as  \( (x + a)^2\) or \( (x – a)^2\) can be expanded as follows:

\( (x + a)^2 = (x + a)(x + a) = x^2 + 2ax +a^2 \)

\( (x – a)^2 = (x – a)(x – a) = x^2 – 2ax +a^2 \)

Methods of finding solutions to Quadratic Equations

A quadratic equation’s root or solution is essentially a number that satisfies the equation. The value on the right side of the equation will be equal to the value on the left side if we substitute the number on the left side.

Any quadratic equation can be solved using one of three methods:

• Factoring approach
• Completing the square method
• Quadratic formula method

The factoring approach is one of the fundamental methods for solving a quadratic equation. However, this approach is only helpful for a particular class of quadratic equations. On the other hand, we can use the quadratic formula and complete the square procedures to resolve almost all varieties of quadratic problems. Moreover, the completing square technique is helpful because not all such equations can be factored.

Completing the square when the coefficient of x is 1

Example: Use completing the square formula to solve: \( x^2 – 6x + 9 = 0 \).

Solution:

Using the formula, \(ax^2 + bx + c = a(x + m)^2 + n\).
Here, a = 1, b = -6, c = 9.

\(=> m = \dfrac{b}{2a} = \dfrac{-6}{2(1)} = -3\)

and, \(n = c – (\dfrac{b^2}{4a}) = 9 – \dfrac{(-6)^2}{4(1)} = 9 – \dfrac{36}{4} = 9 – 9 = 0\)

\(=> x^2 – 6x + 9 = (x – 3)^2\)

\(=> (x – 3)^2 = 0\)

\(=> x – 3 = \pm \sqrt{0}\)

\(=> x – 3 = 0\)

\(=> x = 3\)

After completing the square, the equation simplifies to \(=> (x – 3)^2 = 0\). Since the square of any real number is non-negative, the only solution is when (x – 3) equals zero, leading to x = 3.

Completing the square when the coefficient of x is not 1

Example: Use completing the square formula to solve the equation: \(2x^2 + 6x – 5 = 0\)

Solution:

Using the formula, \(ax^2 + bx + c = a(x + m)^2 + n\)
Here, a = 2, b = 6, c = -5.

Divide the equation by the coefficient of \(x^2\):
\(x^2 + 3x – \dfrac{5}{2} = 0\)

Move the constant term to the other side of the equation:
\(x^2 + 3x = \dfrac{5}{2}\)

Take half of the coefficient of \(x (\frac{3}{2})\) and square it:
\((\dfrac{3}{2})^2 = \dfrac{9}{4}\)

Add the squared value obtained in previous step to both sides of the equation:
\(x^2 + 3x + \dfrac{9}{4} = \dfrac{5}{2} + \dfrac{9}{4}\)

Simplifying:
\((x + \dfrac{3}{2})^2 = \dfrac{19}{4}\)

Take the square root of both sides (considering both the positive and negative square roots):
\(x + \dfrac{3}{2} = \pm\sqrt{\dfrac{19}{4}}\)

Solve for x by subtracting \(\frac{3}{2}\) from both sides:
\(x = -\dfrac{3}{2} \pm \sqrt{\dfrac{19}{4}}\)

Therefore, the solutions to the equation \(2x^2 + 6x – 5 = 0\) after completing the square are
\(x = -\dfrac{3}{2} + \dfrac{\sqrt{19}}{2}\) and \(x = -\dfrac{3}{2} – \dfrac{\sqrt{19}}{2}\)

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