VERIFYING THE RELATION BETWEEN ZEROES AND COEFFICIENTS

The basic relationship between zeroes and coefficient of quadratic equation ax2 + bx+ c  =  0 are 

Sum of zeroes  =  α + β  =  -b/a 

Product of zeroes  =  α β  =  c/a

Here α  and β are zeroes of the quadratic polynomial. a, b and c are the coefficients of x2, x and constant terms respectively.

Question :

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients

(i) x² – 2 x – 8

Solution :

First, let us find the zeroes of the polynomial. 

x² – 2 x – 8 = 0

(x – 4) ( x + 2) = 0

x - 4  =  0

x  =  4

x + 2  =  0

x  =  -2

So, α = 4 and β = -2

Now we are going to verify the relationship between these zeroes and coefficients

x² – 2 x – 8 = 0

ax² + b x + c = 0

a = 1   b = -2   c = -8

Sum of zeroes α + β = -b/a

4 + (-2) = -(-2)/1

2 = 2

Product of zeroes α β = c/a

4(-2) = -8/1

-8 = -8

(ii) 4 s² – 4 s + 1 

Solution :

First, let us find the zeroes of the quadratic polynomial.

4 s² – 4 s + 1

(2 s – 1) ( 2 s - 1) = 0

2 s – 1 = 0    

2 s = 1

s = 1/2,   s = 1/2

So,  α = 1/2 and β = 1/2

Now we are going to verify the relationship between these zeroes and coefficients

4 s² – 4 s + 1 = 0

ax² + b x + c = 0

a = 4   b = -4   c = 1

Sum of zeroes α + β = -b/a

(1/2) + (1/2) = -(-4)/4

2/2 = 1

1 = 1

Product of zeroes α β = c/a

(1/2)(1/2) = 1/4

1/4 = 1/4

(iii)  6 x² – 3 – 7 x 

Solution :

First, let us find the zeroes of the quadratic polynomial.

6 x² – 3 – 7 x = 0

6 x²  - 7 x – 3 = 0

3x + 1  =  0

3x  =  -1

x  =  -1/3

2x - 3  =  0

2x  =  3

x  =  3/2

So, α = -1/3 and β = 3/2

Now we are going to verify the relationship between these zeroes and coefficients

6 x²  - 7 x – 3 = 0

ax² + b x + c = 0

a = 6   b = -7   c = -3

Sum of zeroes α + β = -b/a

(-1/3) + (3/2) = -(-7)/6

(- 2 + 9)/6 = 7/6

7/6 = 7/6

Product of zeroes α β = c/a

(-1/3)(3/2) = -3/6

-1/2 = -1/2

(iv)  4 u² + 8 u 

Solution :

Find let us find the zeroes of the quadratic polynomial.

4 u² + 8 u = 0

4 u (u + 2) = 0

 4 u = 0       u + 2 = 0

   u = 0            u = -2

So, α = 0 and β = -2

Now we are going to verify the relationship between these zeroes and coefficients

4 u² + 8 u = 0

ax² + b x + c = 0

a = 4   b = 8   c = 0

Sum of zeroes α + β = -b/a

0 + (-2) = -8/4

- 2 = -2

Product of zeroes α β = c/a

(0)(-2) = 0/4

0 = 0

(v) t² - 15 

Solution :

Find let us find the zeroes of the quadratic polynomial.

t2 - 15 = 0

t2  = 15

  t = √15

  t = ± √15

t = √15   t = - √15

So the values of α = √15 and β = -√15 

Now we are going to verify the relationship between these zeroes and coefficients

t² - 15 = 0

ax² + b x + c = 0

a = 1   b = 0   c = -15

Sum of zeroes α + β = -b/a

√15 + (-√15) = -0/1

0 = 0

Product of zeroes α β = c/a

(√15)( -√15) = -15/1

-15 = -15

(vi) 3 x² – x - 4 

Solution :

Find let us find the zeroes of the quadratic polynomial.

3 x² – x - 4 = 0

(3 x - 4) (x + 1) = 0

(3 x - 4) = 0           (x + 1) = 0

3 x = 4                   x = -1

   x = 4/3          

x = 4/3   x = -1

So the values of α = 4/3 and β = -1

Now we are going to verify the relationship between these zeroes and coefficients

3 x² – x - 4 = 0

ax² + b x + c = 0

a = 3   b = -1   c = -4

Sum of zeroes α + β = -b/a

(4/3) + (-1)  =  -(-1)/3

(4-3)/3  =  1/3

1/3  =  1/3

Product of zeroes α β = c/a

                            (4/3)( -1)  =  -4/3

                               -4/3  =  -4/3

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Derivative of Absolute Value of x Using Limit Definition

    Apr 23, 25 11:11 AM

    Derivative of Absolute Value of x Using Limit Definition

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 149)

    Apr 23, 25 02:33 AM

    digitalsatmath182.png
    Digital SAT Math Problems and Solutions (Part - 149)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 148)

    Apr 22, 25 08:20 AM

    digitalsatmath180.png
    Digital SAT Math Problems and Solutions (Part - 148)

    Read More