NATURE OF ROOTS OF QUADRATIC EQUATION DISCRIMINANT EXAMPLES

Nature of Roots of Quadratic Equation Discriminant Examples :

The roots of the quadratic equation ax² +bx +c = 0 , a ≠ 0 are found using the formula x = [-b ± √(b² - 4ac)]/2a

Here, b² - 4ac called as the discriminant (which is denoted by D ) of the quadratic equation, decides the nature of roots as follows

Nature of Roots of Quadratic Equation Discriminant Examples

Example 1 :

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

(i) 2 x² - 3 x + 5  =  0

Discriminant  =  b² - 4 a c

a  =  2  b  =  -3 and c  =  5

  =  (-3)² - 4 (2) (5)

  =  9 - 40

  = - 31 < 0

It has no real roots.

(ii) 3 x² - 4 √3 x + 4 = 0

Solution :

Discriminant  =  b² - 4 a c

a  =  3  b  =  - 4 √3 and c  =  4

  =  (- 4 √3)² - 4 (3) (4)

  =  16 (3) - 48

  =  48 - 48

  =  0

It has equal real roots.

 3x² - 2√3 x - 2 √3 x + 4  =  0

 3x (√3 x - 2) + 2 (√3 x - 2)  =  0 

 (3 x + 2) (√3 x - 2)  =  0  

 3 x + 2  =  0          √3 x - 2  =  0

   3 x  =  -2                  √3 x  =  2 

     x  =  -2/3                    x  =  2/√3

(iii) 2 x² - 6 x + 3  =  0

Solution :

Discriminant = b² - 4 a c

a = 2  b = - 6 and c = 3

  =  (-6)² - 4 (2) (3)

  =  36 - 24

  =  12 > 0

It has two distinct real roots.

We cannot factorize the given equation. To solve this we have to use the quadratic formula

  x  =  (- b ± √ b² - 4 a c)/2a

  x  =  [-(-6) ± √12]/2(2)

 x  =  [6 ± √12]/4

  x  =  [6 ± 2√3]/4

  x  =  2 [3 ± √3]/4

 x  =  (3 ± √3)/2

Example 2 :

Find the values of k for which of the following quadratic equations, so that they have two equal roots.

(i) 2 x² + k x + 3 = 0

Solution :

Since the equation has two equal roots

Discriminant = 0

b² - 4 a c = 0

a = 2  b = k and c = 3

 k² - 4 (2) (3)  =  0

k² - 24  =  0

k²  =  24

 k  =  √24

k  =  √2 ⋅ 2 ⋅ 2 ⋅ 3

k  =  2√6 

(ii) k x (x - 2) + 6  =  0

Solution :

k x² - 2 k x + 6  =  0

Since the equation has two equal roots

Discriminant  =  0

b² - 4 a c  =  0

a = k  b = -2 k and c = 6

(-2 k)² - 4 (k) (6)  =  0

4 k² - 24 k  =  0

4 k (k - 6)  =  0

4 k  =  0            k - 6  =  0

k  =  0                k  =  6

After having gone through the stuff given above, we hope that the students would have understood, how to find the nature of roots of quadratic equation using discriminant.

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