Quadratic Formula Calculator - Quadratic Equation Solver (2024)

The Quadratic Formula

\( x = \dfrac{ -b \pm \sqrt{△}}{ 2a } \)

The quadratic equation has only one root when \( △ = 0 \). The solution is equal to \( x = \frac{-b}{2a} \) it is called repeated or double roots.

The quadratic equation has no real solution for \( △ < 0 \).

Notice that as \( △ < 0 \), the square root of the determinant will be an imaginary value

Hence,

\( Re(x) = \dfrac{-b}{2a} \)

\( lm(x) = \pm \dfrac{△}{2a} \)

Quadratic Form

There are three main forms of quadratic equations:
1- Standard form
2- Factored form
3- Vertex form

Roots Of The Quadratic Equation

The values of x satisfying the equations are called the roots of the equation. A second degree equation which contain at least one term that is Squared.

Methods to Solve Quadratic Equations

Quadratic equation can be solved in multiple ways
1- Factoring
2- Using the quadratic formula
3- Completing the square
4- Graphing

How to drive Quadratic Formula

In general, the steps taken to solve a quadratic equation \( x^2 + bx + c = 0 \) where a, b and c are real numbers, by completing the square.

\( ax^2 + bx + c = 0 \)

Divide throughout by “a”:

\( \dfrac{a}{a}x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0 \)

\( x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0 \)

Rewrite the equation such that the constant term is on the RHS of the equation:

\( x^2 + \dfrac{b}{a}x = -\dfrac{c}{a} \)

Add \( (\frac{b}{2a})^2 \) to both sides of the equation to make LHS a perfect square:

\( x^2 + \dfrac{b}{a}x + (\dfrac{b}{2a})^2 = -\dfrac{c}{a} + (\dfrac{b}{2a})^2 \)

Factorise the expression on the LHS and simplify RHS:

\( (x + \dfrac{b}{2a})^2 = -\dfrac{c}{a} + \dfrac{b^2}{4a^2} \)

\( (x + \dfrac{b}{2a})^2 = \dfrac{b^2 – 4ac}{4a^2} \)

Take sqaure root on both sides

\( \sqrt{(x + \dfrac{b}{2a})^2} = \pm \sqrt{\dfrac{b^2 – 4ac}{4a^2}} \)

\( x + \dfrac{b}{2a} = \pm \dfrac{\sqrt{b^2 – 4ac}}{2a} \)

\( x = -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 – 4ac}}{2a} \)

\( x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)

Graphing

The X value found through the quadratic formula are roots of quadratic equation. X values where any parabola cross the x-axis.

Quadratic formula also define the axis of symmetry of the parabola.

Our quadratic equation solver will solve a second degree polynomial equation such as \( ax^2 + bx + c = 0 \) for x, where \( a ≠ 0 \) for real and complex root where x is an unknown, A is referred to as the quadratic co-efficient, B is the linear co-efficient and C Is the constant. By considering discriminant value \( b^2 – 4ac \), we can assess the nature of the roots.
Check our Discriminant Calculator.

When
\( b^2 – 4ac < 0 \) there is no real root.
\( b^2 – 4ac = 0 \) there is one real root.
\( b^2 – 4ac < 0 \) there is no real root.

Note: Quadratic equation are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable we use a quadratic equation.

When we solve quadratic equation by graphing method, here x values are found through the quadratic formula where any parabola crosses the x-axis. Quadratic formula also define the axis of symmetry of the parabola. If any equation that can be rewrite in \( ax^2 + bx + c = 0 \) form can be solved by using quadratic formula.

We can also graph the function \( y = ax^2 + bx + c \). Its shape is parabola and the roots of the quadratic equation are the x-intercepts of this function.

A Random Image Of Cup-Up Parabola

\( y = 2x^2 – 4x – 1 \)

A Random Image Of Cup-Down Parabola

\( x^2 – 2x + 1 = 0 \)

For more details visit Graph Presentation “Parabola” Of A Quadratic Equation

Note: The above mentioned formula for solving quadratic equation is normally used when the quadratic expression cannot be factorized easily.

Note: Another method that can be used to find the solutions of quadratic equation \( ax^2 + bx + c = 0 \) is by drawing the corresponding quadratic graph of \( y = ax^2 + bx + c \) and to find the x–coordinates of the points of intercepts of the graphs with the x-axis (y = 0).

Regardless of which method you are adopting you first need to check whether the quadratic equation should equal to zero.

Examples

Solve the equation \(3x^2 + 4x – 5 = 0 \)

Solution: It is a quadratic equation but it can be solved by factorization so we should opt for quadratic formula or either completing square method first.
For more details visit Quadratic Equation by Factoring Method

By completing the square method

\(3x^2 + 4x – 5 = 0 \)

\(3x^2 + 4x = 5 \)

\(x^2 + \dfrac{4}{3}x = \dfrac{5}{3} \)

\(x^2 + 2(\dfrac{2}{3})(x) + (\dfrac{2}{3})^2 = \dfrac{5}{3} + (\dfrac{2}{3})^2 \)

\(( x + \dfrac{2}{3})^2 = \dfrac{5}{3} + \dfrac{4}{9} \)

\(x + \dfrac{2}{3} = \pm \dfrac{\sqrt{19}}{9} \)

\(x = -2.12 \) and \(x = 0.78 \) are the roots of the equation

By applying quadratic formula

Comparing \(3x^2 + 4x – 5 = 0 \) which \(ax^2 + bx + c = 0 \)

We have \(a = 3\), \(b = 4\) and \(c = 5\)

After substituting the above values in quadratic formula,

\( x = \dfrac{ -b \pm \sqrt{b^2 – 4ac}}{ 2a } \)

Result: \(x = 0.786\) or \(x = -2.12\)

Solve the equation \( y = x^2 – 4x + 4 \)

By graphical method

Complete the table

Draw the graph

Now table will be

From the graph, the curve \( y = x^2 – 4x + 4 \) touches the x-axis at \( x = 2 \) only. Therefore, the solution of the equation \( y = x^2 – 4x + 4 \) is \(x = 2\).

Quadratic Functions And Their Graphs:

Standard form of Quadratic Equation

\( ax^2 + bx + c = 0 \)

This is the quadratic equation in standard form. In this form of quadratic equations we have,

• a, b and c are constants
• a is not equals to zero (\( a ≠ 0 \))

For Example:

Lets suppose three equations A, B and C.

Equation A: \( x^2 + 5x + 2 = 0 \), here \( a = 1 \), \( b = 5 \) and \( c = 2 \)

Equation B: \( 3x^2 + x + 9 = 0 \), here \( a = 3 \), \( b = 1 \) and \( c = 9 \)

Equation C: \( x + 9 = 0 \), here \( a = 0 \), \( b = 1 \) and \( c = 9 \)

Equation A and B are standard form of quadratic equation, but equation C not belongs to standard form of quadratic because in this equation, \( a = 0 \)

Pure/Incomplete Form Of A Quadratic Equation

Any equation of the form \( ax^2 + bx + c = 0 \), in the variable x, here a, b, c are real numbers and \(a ≠ 0\) or the quadratic equation having only second degree variable is called a pure quadratic equation.

\( ax^2 + bx + c = 0 \) is called incomplete. In which, either b or c or both equal to 0. Such equation can be easily solved by applying basic algebraic rule. For example, \( x^2 – 49 = 0 \). Thus, to solve such equation we need to take square root on both the sides so \( x^2 = 49 \) becomes \( x = \pm 7 \). So, a quadratic equation that does not contain a first power variable (x term) is called an incomplete or pure quadratic equation.

Often we encounter/came across an equation that is a quadratic in disguise. For example, \( x^4 + 5x^2 – 6 = 0 \)

It’s a fourth degree equation it means exponent of the leading term is twice the second term. Here we have to tackle it by substitution

Let \(z = x^2\)

Then, \(x^4 = z^2\)

Then the equation in terms of z becomes:

\(z^2 + 5z – 6 = 0\)

Now, it clear reflect quadratic equation form.

Hence, the roots are \(z = -3\) and \(z = 1\).

Since \(z = x^2\) we have that \(x^2 = 1\) or \(x^2 = -3\)

The first has solution \(x = 1\) or \(x = -1\)

And the second has no solution.

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