[Maths Class Notes] on Quadratic Polynomial Formula Pdf for Exam

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. The quadratic polynomial formula to find the solutions of the quadratic equation is:

x = [frac{-b pm sqrt{b^{2}- 4ac}}{2a}]

The symbol of ± 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}]

Standard Form of Quadratic Formula 

The standard form of the quadratic formula is ax2+bx+c.

Uses of Quadratic Formula 

Let us also understand the uses of quadratic equations. Interestingly, it is used in everyday life. For example, to find the speed of an athlete or to measure the areas of a room, to calculate the carpet size, or even to determine the profit or loss of a business. 

Learn the Derivations: 

Practising more derivations assists clear understanding of concepts, which helps in remembering concepts in the long run. Deriving one formula introduces you with many more new formulas with conceptual understanding, and there is no need to cram them later.

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

The 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 [ b^{2} – 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 equation are

[x = -[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 to 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

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 equations for x:

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

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  

x + 2= 0 or 2x -3=  0

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

Determine the nature of the roots of the following quadratic equations. If the real roots exist, 2x² -6x + 3 =0

Solution:

If we 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 are 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