 # ⓘ Quadratic equation. A quadratic equation is an equation in the form of a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0}, where a is not equal to 0. It makes a .. A quadratic equation is an equation in the form of a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0}, where a is not equal to 0. It makes a parabola when graphed on a coordinate plane.

When people work with quadratic equations, one of the most common things they do is to solve it. This means to find the points on a coordinate grid where the graphed equation crosses the x-axis, or the horizontal axis. These points are called the "zeroes", or "roots", of a function. In other words, to solve the equation means to find the value of x so that a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0}, where the coefficients, a, b, and c, are all real numbers. There are different ways to find out the value of x. Plotting the function on a graph is one way. In the picture on the right, the function crosses the x-axis at the points − 1, 0 {\displaystyle -1.0} and 2, 0 {\displaystyle 2.0}.

Graphing a quadratic equation makes it is very easy to find the roots, but it is not always convenient. So there are different ways to do this using maths.

## 1. Factoring

Some quadratic equations, like 2 x 2 − x − 1 = 0 {\displaystyle 2x^{2}-x-1=0}, can be factorized. This equation can be written as 2 x + 1 x − 1 = 0 {\displaystyle 2x+1x-1=0}, which produces two separate equations, 2 x + 1 = 0 {\displaystyle 2x+1=0} and x − 1 = 0 {\displaystyle x-1=0}. This is because when any of the two equations is equal to zero, the original equation is equal to zero because the two equations are multiplied together to form the quadratic equation. From this equation, it is easy to use algebra to find out what the value of x is. In the first of the two equations, x = − 1 / 2 {\displaystyle x=-1/2} and in the second, x = 1 {\displaystyle x=1}. This means that these two values of x will make the whole quadratic equation equal to zero. The y-coordinate is equal to zero at these values of x, and so the coordinates − 1 / 2, 0 {\displaystyle -1/2.0} and 1, 0 {\displaystyle 1.0} are the places where the graph of the function crosses the x-axis.

## 2. Completing the square

Not all quadratic equations can be factorized, such as 5 x 2 − 4 x − 2 = 0 {\displaystyle 5x^{2}-4x-2=0}. One of the ways to solve these kinds of equations is called completing the square. This means to find some value for k below so that: x 2 + b a x + k = x 2 + 2 x y + y 2, {\displaystyle x^{2}+{\frac {b}{a}}x+k=x^{2}+2xy+y^{2},\,\!} for some y. Since y = b 2 a {\displaystyle y={\frac {b}{2a}}\,\!} and k = y 2, {\displaystyle k=y^{2},\,\!} this means k = b 2 4 a 2. {\displaystyle k={\frac {b^{2}}{4a^{2}}}.\,\!}

There are five steps to solve a quadratic equation by completing the square:

• Divide the equation by the value of a here, a is equal to 5: x 2 − 4 5 x − 2 5 = 0 {\displaystyle x^{2}-{\frac {4}{5}}x-{\frac {2}{5}}=0}
• Move the number term the one without any coefficients, here 2 5 {\displaystyle {\frac {2}{5}}} to the other side: x 2 − 4 5 x = 2 5 {\displaystyle x^{2}-{\frac {4}{5}}x={\frac {2}{5}}}
• Complete the square by adding k = b 2 4 a 2 {\displaystyle k={\frac {b^{2}}{4a^{2}}}} to both sides on the equation, was described above. This value here is k = 16 100 = 4 25 {\displaystyle k={\frac {16}{100}}={\frac {4}{25}}}. This gives x 2 − 4 5 x + 4 25 = 10 25 + 4 25 {\displaystyle x^{2}-{\frac {4}{5}}x+{\frac {4}{25}}={\frac {10}{25}}+{\frac {4}{25}}}, and the left side can be factored to give x − 2 5 2 = 14 25 {\displaystyle \leftx-{\frac {2}{5}}\right^{2}={\frac {14}{25}}}
• Find the square root of both sides. The ± {\displaystyle \pm } is there since the square root is for both positive and negative values: x − 2 5 = ± 14 25 {\displaystyle x-{\frac {2}{5}}=\pm {\sqrt {\frac {14}{25}}}}
• Finally, move the number term to the other side to get a value for x: x = ± 14 25 + 2 5 {\displaystyle x=\pm {\sqrt {\frac {14}{25}}}+{\frac {2}{5}}}

This last value for x has two values: x = 14 25 + 2 5 ≈ 1.148 {\displaystyle x={\sqrt {\frac {14}{25}}}+{\frac {2}{5}}\approx 1.148} and x = − 14 25 + 2 5 ≈ − 0.348 {\displaystyle x=-{\sqrt {\frac {14}{25}}}+{\frac {2}{5}}\approx -0.348}. These are the x-coordinates at the points where the y-coordinates are equal to zero and so the function crosses the x-axis.

It is also interesting to note that the vertex of the function, 2 5, − 14 25 {\displaystyle \left{\frac {2}{5}} {\frac {14}{25}}\right}, can be found from the equation in step 3 above.

Although completing the square can be used to solve many quadratic equations, it does not work with every equation. There is a way to find the value of x that makes the equation true by replacing the values in a formula with the coefficients from the quadratic equation. This is called the "quadratic formula", which is:

x = − b ± b 2 − 4 a c 2 a {\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}

The numbers represented by a, b, and c are the same as those in the original equation, ax 2 + bx + c = 0. For this formula to work properly, a cannot be equal to 0 for the formula.

The factored form of this equation is y = a x − sx − t, where s and t are the zeros, a is a constant, and y and the two values of x are ordered pairs which satisfy the equation.

### 3.1. The quadratic formula Proof of the formula

The quadratic formula can be found out, or proved, using the same completing the square method from above, only by using a, b, and c instead of actual values and using algebra to find the value of x. Here is how:

• Finally, move b / 2 a to the other side: x = − b 2 a ± b 2 − 4 a c 2 a = − b ± b 2 − 4 a c 2 a {\displaystyle x=-{\frac {b}{2a}}\pm {\frac {\sqrt {b^{2}-4ac\ }}{2a}}={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}}
• Divide the equation by a: x 2 + b a x + c a = 0 {\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0}
• Find the square root of both sides: x + b 2 a = ± b 2 − 4 a c 2 a {\displaystyle x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac\ }}{2a}}}
• Complete the square by adding k = b 2 4 a 2 {\displaystyle k={\frac {b^{2}}{4a^{2}}}} to both sides on the equation, which gives: x 2 + b a x + b 2 4 a 2 = − c a + b 2 4 a 2. {\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {b^{2}}{4a^{2}}}=-{\frac {c}{a}}+{\frac {b^{2}}{4a^{2}}}.\,\!} The left side is now a perfect square; it is the square of x + b 2 a. {\displaystyle x+{\frac {b}{2a}}.\,\!} The right side can be a single fraction, with a common denominator 4 a 2: x + b 2 a 2 = b 2 − 4 a c 4 a 2 {\displaystyle \leftx+{\frac {b}{2a}}\right^{2}={\frac {b^{2}-4ac}{4a^{2}}}}
• Move c / a to the other side: x 2 + b a x = − c a {\displaystyle x^{2}+{\frac {b}{a}}x=-{\frac {c}{a}}}