# Where does the quadratic formula come from?

**Algebra** is a branch of mathematical study that deals with the relation between constants and variables. Constants stand for the known definite values whereas, variables stand for unknown values. In algebra basic operations like summation, subtraction, multiplication, and divisions are used to gain a solution for these unknown values.

The given article is also about the study of solving algebraic expressions using the quadratic formula. The given article is composed with the description of the quadratic formula, its history, derivation method along with some sample problems with their solutions for better understanding.

### Where does the quadratic formula come from?

Quadratic equations are the second-degree algebraic expressions that are written in the standard form of** ax ^{2}+bx+c=0 **where, a,b and c are the known constant values and a is not equal to zero and x is the variable that represents the unknown value in the equation.

The standard quadratic formula is given as

x = -b±√(b^{2}-4ac)/2a

where, a,b and c are the constants.

x is the variable

and, a≠0

**Different conditions for discriminant (b ^{2}-4ac)**

- Condition 1. When the discriminant (b
^{2}-4ac) is positive.

=>There will be two solutions or values derived for the variable.

- Condition 2. When the discriminant (b
^{2}-4ac) is zero.

=>There will be only one solution.

- Condition 3. When the discriminant (b
^{2}-4ac) is negative.

=>There will be a pair of complex numbers as solutions.

**History of Quadratic Equation**

- Initially, quadratic equations were solved by geometric methods.
- The Egyptian Berlin Papyrus(2050-1650BC) contains the solution to a two-term quadratic equation.
- The Greek mathematician Diophantus solved quadratic equations with an algebraic method which was more recognizable than the geometric method.
- The Indian mathematician Brahmagupta has described the quadratic formula in his treatises written in words instead of symbols.

**Derivation of quadratic square root formula**

Consider a second degree quadratic equation ax^{2}+bx+c=0.

where a≠0

Now, to determine the roots of this equation

=>ax

^{2}+bx=-cDividing both sides by ‘a’

=>x

^{2}+bx/a=-c/aAdding a new term (b/2a)

^{2}on both sides=>x

^{2}+ bx/a+ (b/2a)^{2}=-c/a +(b/2a)^{2}As the left hand side is now a perfect square.

=>(x+b/2a)

^{2}=-c/a + b^{2}/4a^{2}=>(x+b/2a)

^{2}= (b^{2}-4ac)/4a^{2}Now, we can take square roots to obtain

=> x+b/2a=±√(b

^{2}-4ac)/2a=>x=-b±√(b

^{2}-4ac)/2aThus, by computing the squares we will obtain two roots of the equation.

### Sample Questions

**Question 1. Solve the equation 3x ^{2}+5x-7=0, using the quadratic formula.**

**Solution:**

The given equation is 3x

^{2}+5x-7=0a = 3

b = 5

c = -7

Now,

=>x = -b±√(b

^{2}-4ac)/2a=>x = -b±√(5)

^{2}-4(3)(-7)/2(3)=>x = -5±√25+84/6

=>x = -5±√109/6

Then,

x = -5+√109/6 = 0.907

x = -5-√109/6 = -2.573

**Question 2. Solve the equation x ^{2}+3x+2=0.**

**Solution:**

The given equation is x

^{2}+3x+2=0a=1

b=3

c=-2

Now,

=>x=-b±√(b

^{2}-4ac)/2a=>x=-3±√(3)

^{2}-4(1)(-2)/2(1)=>x=-3±√9-8/2

=>x=-3±1/2

Then,

=>x=-3+1/2=-1

=>x=-3-1/2=-2

**Question 3. Solve the equation x ^{2}+6x+8.**

**Solution:**

The given equation x

^{2}+6x+8a=1

b=6

c=8

Now,

=>x=-b±√(b

^{2}-4ac)/2a=>x=-6±√(6)

^{2}-4(1)(8)/2(1)=>x=-6±√36-32/2

=>x=-6±2/2

Then,

=>x=-6+2/2=-2

=>x=-6-2/2=-4