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CIE A-Level Maths Study Notes

1.1.2 The Discriminant

The discriminant is a pivotal concept in the study of quadratic equations. It not only reveals the nature of the roots but also their quantity, thus enriching our understanding of these equations.

Discriminant of a Quadratic Expression

The discriminant relates to the roots of a quadratic equation in standard form ax2+bx+c=0.ax^2 + bx + c = 0.

Definition:

The discriminant of the quadratic equation ax2+bx+cax^2 + bx + c is denoted and calculated as: Δ=b24ac \Delta = b^2 - 4ac

Interpretation of the Discriminant

The value of the discriminant provides crucial information about the roots of the quadratic equation:

  • Positive Discriminant ( \Delta > 0 ):
    • Indicates two distinct real roots.
    • The roots are real and different because the square root of a positive number is real and distinct.

positive discriminant

Image courtesy of Mathsathome

  • Zero Discriminant (Δ=0)( \Delta = 0 ):
    • Indicates one real root, repeated twice.
    • This situation arises because the square root of zero is zero, leading to a single, repeated solution.

zero discriminant

Image courtesy of Mathsathome

  • Negative Discriminant ( \Delta < 0 ):
    • Indicates no real roots.
    • The square root of a negative number is not real, hence no real solutions exist.

negative discriminant

Image courtesy of Mathsathome

Example 1:

For the equation 3x26x+2=0 3x^2 - 6x + 2 = 0, determine the nature of the roots.

Solution:

Δ=(6)24×3×2\Delta = (-6)^2 - 4 \times 3 \times 2 =3624= 36 - 24=12=12

Since \Delta > 0, the equation has two distinct real roots.

Examples 2:

Find the condition for mm in the equation x2(m+3)x+m=0x^2 - (m + 3)x + m = 0 to have real and equal roots.

Solution:

Δ=((m+3))24×1×m \Delta = (-(m + 3))^2 - 4 \times 1 \times m \ =(m+3)24m= (m + 3)^2 - 4m=m2+6m+94m= m^2 + 6m + 9 - 4m=m2+2m+9= m^2 + 2m + 9

For real, equal roots, Δ=0 \Delta = 0 \

m2+2m+9=0m^2 + 2m + 9 = 0

Thus, mm must satisfy the equation m2+2m+9=0m^2 + 2m + 9 = 0 for the roots to be real and equal.

Example 3:

Determine whether the equation 5x2+2x+1=05x^2 + 2x + 1 = 0 has any real roots.

Solution:

Δ=224×5×1 \Delta = 2^2 - 4 \times 5 \times 1 \ =420= 4 - 20 =16= -16

Since \Delta < 0, the equation has no real roots.

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