RESULTS
What is the quadratic formula?
The quadratic formula is one of the three main ways to find the roots of a second-degree polynomial equation, others are completing square and breaking into factors. Quadratic formula is
\(x = \dfrac{-b\plusmn\sqrt{b^2-4ac}}{2a}\)
Where
\(b2- 4ac\) is also called the discriminant of the equation represented by delta.
What is the quadratic equation?
A second-degree polynomial equation is one in which the highest power of variable (e.g x) is 2. For example, \(4x^2-22x = x^2+5, -2x^2+x+5=0, -2x^2+15=x^2-12\). For the implication of the quadratic formula on the quadratic equation, the equation must be solved to \(ax^2+bx+c=0\) form. For instance, let an equation be \(5x^2-10x+1=x+4\). if we want to find the roots through the quadratic formula, we will have to get one coefficient for one distinct power of the variable i.e. \(5x^2-11x-3=0\).
Discriminant:
Discriminant provides important information about the roots of the equation.
- If the discriminant is greater than zero then after solving the quadratic equation formula, we get two real roots.
- If the discriminant is equal to zero, we get a perfect square as a root.
- If the discriminant is less than zero, both of the roots of the equation are imaginary.
Real roots:
When coefficients a,b and c are real numbers and have no imaginary part i.e i. Also \(a \neq 0\). E.g consider a quadratic equation \(x^2+5x + 6\). if we solve it through quadratic formula, the steps are following
- \(a = 1, b = 5\) and \(c = 6\).
- Putting in formula \(x_1 =\dfrac{-5\sqrt{+5^2-4(1)(6)}}{2(1)}\) and \(x_2 = \dfrac{-5\sqrt{-52-4(1)(6)}}{2(1)}\)
- By solving we get \(x_1=-2\) and \(x_2 =-3\).
- Roots are real
Imaginary roots:
Roots having an imaginary number in them. The simplest imaginary number is \(i = -1\). Such roots do have real numbers in them but due to the presence of i, the whole root is called imaginary. E.g let us solve the equation \(x^2-5x +7\).
- \(a = 1, b = -5\) and \(c = 7\)
- Putting in formula \(x_1 =\dfrac{-(-5)+\sqrt{(-5)2-4(1)(7)}}{2(1)}\) and \(x_2 =\dfrac{-(-5)-\sqrt{(-5)2-4(1)(7)}}{2(1)}\)
- By solving we get \(x_1= \dfrac{5+\sqrt{-3}}{2}\) and \(x_2= \dfrac{5-\sqrt{-3}}{2}\), where \(\sqrt{-3}\) is 3i.
- Roots are imaginary.
Coefficients and variables of a quadratic equation:
The real numbers along with the variable in an equation are called coefficients of a quadratic equation. Coefficients are not dependent on the variable such as x, y, and z. It is not essential that only x can be the variable. We can use any alphabet as a variable. Only one type of variable is present in a quadratic equation. It is not possible for a quadratic equation to have more than one variable. If there is no coefficient written with “x”, it means its coefficient is 1. If its second-degree polynomial and coefficient of x2is zero, It’s not a quadratic equation, it is a linear equation.
Derivation of the quadratic formula
The quadratic formula is obtained by completing the square method.
- Take the equation \(ax^2+bx+c=0\).
- Make the coefficient of x^{2} be 1 by dividing the whole equation by a, i.e
\(x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0\). - Add and subtract a suitable square to complete the formula of \((a+b)^2\).i.e
\(x^2+\dfrac{b}{a}x+\dfrac{c}{a}+\Big(\dfrac{b}{2a}\Big)^2-(\Big(\dfrac{b}{2a}\Big)^2=0\)
- \(x^2+\dfrac{b}{a}x+\Big(\dfrac{b}{2a}\Big)^2 = \Big(\dfrac{b}{2a}\Big)^2 - \dfrac{c}{a}\)
- \(\Big(x+\dfrac{b}{2a}\Big)^2 = \Big(\dfrac{b}{2a}\Big)^2 - \dfrac{c}{a}\)
- Taking square root on both sides \(x + \dfrac{b}{2a} = \sqrt{\Big(\dfrac{b}{2a}2\Big) - \dfrac{c}{a}}\)
- \(x = - \dfrac{b}{2a} \plusmn \sqrt{\Big(\dfrac{b}{2a}2\Big) - \dfrac{c}{a}}\)
- \(-\pm\sqrt{-\dfrac{c}{a}\cdot\left(2a\right)^2+\left(\dfrac{b}{2a}\right)^2\cdot\left(2a\right)^2}\)
- \(x = \dfrac{-b\pm\sqrt{-4ac+b^2}}{2a}\)
How to solve the quadratic equation manually?
The quadratic formula is one of the easiest ways to solve a quadratic equation. You can solve a quadratic equation as instructed.
- First, simplify your equation into the basic quadratic form. E.g we have to convert \(4x^2+5x-7=4+x\) into \(4x^2+4x-11=0\).
- Break it into parts a,b and c for the degree of variable 2,1 and 0 respectively. E.g \(a = 4, b = 4, c = -11\)
- Put in the quadratic formula. E.g \(\dfrac{-4\pm\sqrt{4^2-4\left(4\right)-11}}{2\left(4\right)}\)
- First, solve with + sign and secondly, with - sign to get two roots. E.g \(x_1=\dfrac{-1+2\sqrt{3}}{2}\) and \(x_2=\dfrac{-1-2\sqrt{3}}{2}\)
- Note whether the roots are real or imaginary. E.g roots are real and unequal.
How to use the Quadratic formula calculator?
As you can see from the procedure written above for solving the quadratic equation, this work can take a great deal of your time. Why not use a quadratic equation solver which will not only give you answers but also tell you the procedure, nature of roots and give you a plotted graph. Seems impossible? But not for us. Yes! By using our quadratic formula calculator you will not only get the answers but also previously mentioned things. You’ll need to follow only one step to get your answer and that step is
- Write the coefficients of the variable in the text boxes and click calculate.