Quadratic Formula

A quadratic equation, therefore, refers to a mathematical relationship between a variable and the constants, where the highest power of the variable is squared. Sometimes, solving quadratic equations can be difficult, especially when factoring does not work or is not done. One of the best methods available for their solution is the quadratic formula, the famous formula which gives the solutions directly. 

1.0General Form of Quadratic Equation

The General Form of Quadratic Equation is an important aspect of solving quadratic equations as it allows systematic methods for finding the solutions of the equation. The standard form of quadratic equation can be written as: 

$a{x}^2+bx+c=0$

Here, 

• x is the variable or unknown we are solving for, 
• a, b, and c are constants known as the coefficients, and 
• $a\ne 0$ (if a = 0, the equation would be linear, not quadratic, because the highest power of x would then be 1)

2.0Quadratic Equation Formula

The quadratic equation formula, also known as the quadratic formula, is used for finding the roots of the equation in a standard form, which is $(a{x}^2+bx+c=0)$. It provides solutions or roots of the equation without a need to use other tougher methods. The Quadratic Formula can be expressed as: 

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, “$(b^2-4ac)$” is known as the discriminant, which can be symbolised as “D”. It helps in determining the nature of roots: 

• If D＞0, then the roots of the equation will have two real “distinct” roots.
• If D=0, then the equation will have two “equal” roots. 
• If D＜0, then the roots of the equation will be imaginary. 

3.0How to Solve Quadratic Equations by Quadratic Formula 

Solving quadratic equations can become easy by using the following four steps: 

• Step 1: Identify the values of a, b, and c from the quadratic equation.
• Step 2: Put these values into the quadratic formula.
• Step 3: Reduce the expression under the square root and determine whether the discriminant is positive, zero, or negative. The two former cases translate to two real solutions and one real solution, respectively. The latter gives complex solutions.
• Step 4: Solve for x. 

4.0Derivation of Quadratic Equation

The formula for the quadratic equation can be derived by using the standard form of the quadratic equation. Like this: 

$a{x}^2+bx+c=0$

Divide the equation by the “a”

$x^2+\left(\frac{b}{a}\right)x+\left(\frac{c}{a}\right)=0$ …..(1)

Using the completing the square method, taking the coefficient of x, dividing it by ½, and then squaring it: 

$\frac{b}{a}\Rightarrow \frac{b}{2a}\Rightarrow \frac{b^2}{4a^2}$

Adding and subtracting $\frac{b^2}{4a^2}$ on both sides in equation (1) 

$x^2+\left(\frac{b}{a}\right)x+\frac{b^2}{4a^2}+\left(\frac{c}{a}\right)-\frac{b^2}{4a^2}=0$

${\left(x+\frac{b}{2a}\right)}^2+\frac{c}{a}-\frac{b^2}{4a^2}=0$

${\left(x+\frac{b}{2a}\right)}^2+\frac{4ac-b^2}{4a^2}=0$

${\left(x+\frac{b}{2a}\right)}^2=\frac{-(4ac-b^2)}{4a^2}$

$\left(x+\frac{b}{2a}\right)=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

5.0Graphing Quadratic Equations

Graphing quadratic equations is a graphical method of finding the solutions, or roots, of a quadratic equation and understanding the behaviour of the equation written in the standard form $(a{x}^2+bx+c=0)$. This equation produces a parabola, which is a symmetrical curve. The shape of the parabola and its orientation (whether it opens upwards or downwards) depends on the value of a.

Steps for Graphing Quadratic Equations

1. Write the quadratic equation in its standard form.
2. Find the vertex of x - coordinate by using the formula $(x=\frac{-b}{2a})$ and substituting the values of a and b. To find the y coordinate, simply put this value of the x-coordinate in the quadratic equation. 
3. Find the axis of symmetry, which is a vertical passing through the vertex by using again the formula of $(x=\frac{-b}{2a})$
4. Evaluate y when x = 0 and the y-intercept is (0, c) to find the y-intercept. 
5. Draw the parabola by smoothly connecting the points, ensuring the curve opens upward (if a > 0) or downward (if a < 0).
6. The point where the parabola intercepts on the x-axis is the root of the quadratic equations. 

6.0Quadratic Equation Examples

Problem 1: Find the roots of x² + 6x − 7 = 0 using the quadratic formula. 

Solution: Given that, x² + 6x − 7 = 0 

a = 1, b = 6, c = –7

$D=b^2-4ac=6^2-4\times 1\times (-7)$

$D=36+28=64$

$x=\frac{-6\pm \sqrt{64}}{2(1)}=\frac{-6\pm 8}{2}$

$x=\frac{-6+8}{2},x=\frac{-6-8}{2}$

$x=1,-7$

Problem 2: Determine if the given equation x² + 2x + 5 = 0 has a real root or not, if the roots are real, then find the roots of the equation. 

Solution: Given that x² + 2x + 5 = 0

a = 1, b = 2, c = 5 

$D=b^2-4ac=2^2-4(1\times 5)=4-20$

D = –16

Hence, the root of the given equation is imaginary. 

Problem 3: The roots of the following equation x² − bx + 9 = 0 are equal, which is 3. Find the value of b.

Solution: Given that x² − bx + 9 = 0

a = 1, c = 9, b = ?

As the roots of the equations are equal hence, D = 0 

$D=b^2-4ac$

$0=b^2-4\times 1\times 9$

$b^2=36$

$b=\sqrt{36}=6$