Quadratic Polynomial Formula

Quadratic Formula

The quadratic formula is a formula that enables us to find the solutions of quadratic equations. The standard form of the quadratic equation is ax² + bx + c, where a,b and c are real numbers and are also known as numeric coefficients. Here the variable ‘x’ is unknown and we have to find the solution for x. Quadratic polynomial formula to find the solutions of the quadratic equation is 


𝓧 = \[\frac{-b\: \pm\: \sqrt{b^{2} - 4ac }}{2a}\]


The symbol of \[\pm\]  indicates that there are two solutions of quadratic equations such as,


𝓧 = \[\frac{-b\: +\: \sqrt{b^{2}- 4ac}}{2a}\] and 𝓧 = \[\frac{-b\: -\: \sqrt{b^{2}- 4ac}}{2a}\]


Quadratic Polynomial 

A polynomial in the form of ax² + bx + c where a, b and c are real numbers, and a 0 is known as  a quadratic polynomial.

Quadratic Equation

We get a quadratic equation when we equate a quadratic polynomial to a constant 


Any equation represented in the form of p(x)=c, where p(x) is denoted as a polynomial of degree 2 and c is denoted as a constant, is considered as a  quadratic equation.

The standard form of a Quadratic Equation

The standard form of a quadratic equation is ax² + bx + c=0, where ax² + bx + c where a,b and c are real numbers and a0

As, ‘a’ is the coefficient of x². It is known as the quadratic coefficient

‘b’ is the coefficient of x. Hence, it is known as the linear coefficient and 

‘c’ is the constant term.

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Quadratic Polynomial  Formula Example


Let us take the equation x² - 3x + 2= 0


Solution- Substituting x =1 in the LHS gives 

1² - 3(1) + 2

1-3+2 = 0

Hence, LHS = RHS


Now, we will substitute the value of x =2 in the LHS. It will give us

2² - 3(2) + 2

4- 6+ 2

-2+2

= 0

Hence, LHS = RHS


Here, the values of x =1 and x = 2 satisfy the equation x² - 3x + 2 = 0. These are known as solutions or roots of the quadratic equation. It also implies that numbers 1 and 2 are the zeros of the polynomial  x² - 3x + 2.

Quadratic Formula Proof

Examine the equation x² - 3x + 2 = 0. Dividing the LHS of the equation with ‘a’ gives us

x²/a - 3x/a + 2 /a = 0


By using the completing the square method we get


(x +b/2a)² - (b/2a)² + c/a = 0

(x +b/2a)² - (b² - 4ac)/ 4a² = 0

(x +b/2a)² =  (b² - 4ac)/ 4a²


Roots of the quadratic equation are calculated by taking the square root of RHS. Here the value of b² - 4ac should be greater than or equal to 0.


Solving for b² - 4ac > 0


(x + b/2a) = \[\pm\] \[\frac{-b\: -\: \sqrt{b^{2}- 4ac}}{2a}\]


𝓧  = -b \[\pm\] \[\frac{-b\: -\: \sqrt{b^{2}- 4ac}}{2a}\]


Hence, the roots of the equations are


𝓧  = -b + \[\frac{-b\: -\: \sqrt{b^{2}- 4ac}}{2a}\] and x =  -b - \[\frac{-b\: -\: \sqrt{b^{2}- 4ac}}{2a}\]


The quadratic equation will have no real roots if  b² - 4ac < 0 because square roots cannot be defined for the negative numbers in the real number system.


Hence, ax² + bx + c is the quadratic formula to find the roots of the quadratic equation.

How to Split The Middle Term of A Quadratic Equation

To split the middle term of the quadratic equation, our aim is to write quadratic expressions as the product of two- line quadratic polynomials. The method for splitting the middle term is shown in detail , in which we will factor examples of quadratic expressions.

For example, you have a quadratic polynomial that can be expressed as


(Ax + B) (Cx + D)


Let us multiply the above equation and simplify the expression


ACx² + (AD + BC)x + BD


So, in terms of  ax² + bx + c, we have

a = AC

b = AD + BC

c = BD


We will solve the quadratic equations examples from the quadratic expression given below

ACx² + (AD + BC)x + BD


And we make an effort to factor it back into the form

(Ax + B) (Cx + D)


The aim is to find a combination of factors of ABCD that sum up to b = AD + BC.

Quadratic Factorization Using The Splitting of The Middle Term Example


  1. Find the factors of  12x²  - 15 = 11x

Solution: 12x²  - 15 = 11x

 12x² + 11x - 15 =0

 12x² + 20x – 9x - 15 =0

4x ( 3x + 5) -3(3x + 5)= 0

(3x + 5)(4x-3) = 0

3x + 5 = 0 or 4x-3 = 0

3x = -5 or 4x = 3

x = -5/3 or x =3/4

Hence, the solution is (-5/3), (3/4)

Solved Examples

Solve the following quadratic equation for x.

  1. x² - 3x + 10 = 0


Let us express -3x as a sum of -5x and 2x.

  • x² - 5x + 2x -10=0

  • x ( x-5) + 2(x-5) = 0

  • (x-5) ( x+2) = 0

  • x-5 = 0 or x + 2= 0

  • x =5 or x= -2


  1. 2x²  + x- 6

Solution: 2x² + 4x-3x -6

2x ( x + 2) -3( x+ 2)

= (x+2) ( 2x-3)

Roots of this equation are the values for which for  

x + 2= 0 or 2x -3=  0

i.e , x = -2 or x= 3/2


  1. Determine the nature of the roots of the following quadratic equations. If the real roots exist, 

2x² -6x + 3 =0

Solution:

If we will compare the above quadratic equation  with ax² + bx + c= 0, we get

a = 2, b= -6 and c= 3

We know that

D = b² - 4ac

= (-6)² - 4 * 2* 3

= (-6)² * (-6)²  - 4 * 2* 3= 36-24

= 12

As D > 0

There are two different roots

Now using quadratic formula to find roots,

x = -b  \[\pm\]   \[\sqrt{D}\]/ 2a


Substituting values,


x = -(-6) \[\pm\] \[\sqrt{12}\]/2  × 2

x = + 6 \[\pm\] \[\sqrt{12}\]/4

x = 6 \[\pm\] \[\sqrt{4 X 3}\]/4 

x = 6 \[\pm\] \[\sqrt{4}\]   X \[\sqrt{3}\]/4 

Quiz time

  1. The quadratic equations whose roots are -2 and 4 is derived by,

  1. x² - 2x - 8 = 0

  2. x² -2x +8 = 0

  3. x² + 2x - 8 = 0

  4. None of these


2.  The equation whose roots are 1 and 0 is

  1. x² - 2x + 1= 0

  2. x² - 1= 0

  3. x² - x = 0

  4. None of these 


3. The expression - x² + 2x + 1 is always

  1. Positive 

  2. Negative

  3. 0

  4. None of these

FAQ (Frequently Asked Questions)

1. What are the different applications of quadratic equations in real-life?

Some applications of quadratic equations in real-life are :


  1. Learning

The quadratic equation plays an important part in the education system. Thousands of students resolve quadratic equations every day. If you are a Mathematician, physics student or teacher, you must be using these types of calculations each day. 


  1. Sports

The quadratic equation is used in sports very often. It is helpful in anlaysing the scores of the player as well as in gameplay. For instance, when a football analyst wants to determine the form of a team or athlete they always prefer to make calculations. You will find one or two elements of the quadratic equation in the analysis. Basketball players throw the ball into the net and measure the accurate distance and time that it will take. The height of the ball can be calculated using a  velocity quadratic equation


  1. Military or law enforcement

Quadratic equations are used in military or law enforcement to estimate the speed of the moving objects such as cars and planes. The military used quadratic equations to determine the distance between them and an approaching enemy. The traffic police use quadratic equations to determine the speed of the car involved in a car accident on the road.


Apart from the above applications, quadratic equations are also used in 

  • A satellite dish

  • Finding a sped

  • Calculating room areas

  • Engineering

  • Management and clerical work

  • Agriculture

2. What are the roots of quadratic equations?

We know that a second degree polynomial will have a maximum of 2 zeros. Hence, a quadratic equation will have a maximum of two roots.

In general, if α is a root of the quadratic equation ax² + bx + c 0, then aα² + bα + c 0.

We can also say that x = α is a solution of quadratic equation or αagrees satisfy the equation ax ax² + bx + c = 0.

Notes:

  • Roots of the quadratic equation ax² + bx + c = 0 are similar as zeo of the polynomial ax² + bx + c .

  • We can factorize the quadratic polynomial by splitting the middle term.


For example, x² + 5x + 6 can be written as x² + 2x + 3x + 6,= x(x + 2) + 3( x + 2)= (x + 2)(x + 3)


x = 6 ± 2√3/4     

x = 2 (3 ±√3)/4

x = 3 ±√3/ 2


Hence, the root of the equations are x = 3 + √3/2 and  x = 3 - √3/2