NATURE OF ROOTS OF QUADRATIC EQUATION DISCRIMINANT

Nature of Roots of Quadratic Equation Discriminant :

Here we are going to see some example problems  of finding nature of roots of a quadratic equation.

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

Value of discriminant

Δ = b2 - 4ac

Δ > 0

Δ = 0

Δ < 0

Nature of roots


Real and unequal roots

Real and equal roots

No real roots

Nature of Roots of Quadratic Equation Discriminant - Questions

Question 1 :

Determine the nature of the roots for the following quadratic equations

(i) 15x2 + 11x + 2 = 0

Solution :

By comparing the given quadratic equation with the general form of quadratic equation.

ax2 + bx + c = 0

a = 15, b = 11 and c = 2

Δ = b2 - 4ac

Δ = 112 - 4(15)(2)

  Δ  =  121 - 120

  Δ   =  1 > 0

Hence the roots are real and unequal.

(ii) x2 − x − 1 = 0

Solution :

a = 1, b = -1 and c = -1

Δ = b2 - 4ac

Δ = (-1)2 - 4(1)(-1)

  Δ  =  1 + 4

  Δ   =  5 > 0

Hence the roots are real and unequal.

(iii)  √2t2 −3t + 32 = 0

Solution :

a = √2, b = -3 and c = 32

Δ = b2 - 4ac

Δ = (-3)2 - 4(2)(32)

  Δ  =  9 - 12(2)

  Δ  =  9 - 24

  Δ   =  -15 < 0

Hence it has no real roots.

(iv)  9y2 − 6√2 + 2 = 0

Solution :

a = 9, b = − 6√2 and c = 2

Δ = b2 - 4ac

Δ = (− 6√2)2 - 4(9)(2)

  Δ  =  36(2) - 72

  Δ  =  72 - 72

  Δ   =  0

Hence it has no real and equal roots.

(v)  9a2b2x2 −24abcdx + 16c2d2 = 0 , a  0 , b ≠ 0

Solution :

a = 9a2b2, b = −24abcd and c = 16c2d2

Δ = b2 - 4ac

Δ = ( −24abcd)2 - 4(9a2b2)(16c2d2)

  Δ  =  576 a2b2c2d2 576 a2b2c2d2

  Δ  =  0

Hence it has no real and equal roots.

After having gone through the stuff given above, we hope that the students would have understood, "Nature of Roots of Quadratic Equation Discriminant". 

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