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Quadratic Equations - Nature of Roots

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Standard Form of a Quadratic Equation: A quadratic equation is of the form ax2+bx+c=0ax^2 + bx + c = 0, where a,b,a, b, and cc are real numbers and a≠0a \neq 0. The degree of the equation is 2.

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The Discriminant: The nature of the roots depends on the value of the expression b2−4acb^2 - 4ac, which is called the discriminant, denoted by DD. This value determines how many real solutions exist for the equation.

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Case 1 - Two Distinct Real Roots: If D=b2−4ac>0D = b^2 - 4ac > 0, the equation has two distinct real roots. Visually, if you plot the quadratic function y=ax2+bx+cy = ax^2 + bx + c, the parabola will intersect the x-axis at two different points.

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Case 2 - Two Equal Real Roots: If D=b2−4ac=0D = b^2 - 4ac = 0, the equation has two equal real roots (also called coincident roots). Visually, the vertex of the parabola touches the x-axis at exactly one point, meaning the x-axis is a tangent to the curve.

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Case 3 - No Real Roots: If D=b2−4ac<0D = b^2 - 4ac < 0, the equation has no real roots (the roots are imaginary). Visually, the parabola is located entirely above or entirely below the x-axis and never touches or crosses it.

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Relationship with Coefficients: If a,b,ca, b, c are rational and DD is a perfect square, the roots are rational and distinct. If DD is not a perfect square, the roots are irrational and occur in conjugate pairs (e.g., 2+32 + \sqrt{3} and 2−32 - \sqrt{3}).

📐Formulae

Standard Equation: ax2+bx+c=0ax^2 + bx + c = 0

Discriminant: D=b2−4acD = b^2 - 4ac

Quadratic Formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Condition for Real Roots: D≥0D \geq 0

Roots when D=0D=0: x=−b2a,−b2ax = -\frac{b}{2a}, -\frac{b}{2a}

💡Examples

Problem 1:

Determine the nature of the roots of the quadratic equation 2x2−4x+3=02x^2 - 4x + 3 = 0.

Solution:

  1. Identify the coefficients: a=2,b=−4,c=3a = 2, b = -4, c = 3.
  2. Calculate the discriminant: D=b2−4acD = b^2 - 4ac.
  3. Substitute the values: D=(−4)2−4(2)(3)D = (-4)^2 - 4(2)(3).
  4. Solve: D=16−24=−8D = 16 - 24 = -8.
  5. Since D<0D < 0, the equation has no real roots.

Explanation:

To find the nature of roots, we calculate the discriminant. Because the result is a negative number, the roots do not exist in the set of real numbers.

Problem 2:

Find the value of kk for which the quadratic equation kx(x−2)+6=0kx(x - 2) + 6 = 0 has two equal roots.

Solution:

  1. Rewrite the equation in standard form ax2+bx+c=0ax^2 + bx + c = 0: kx2−2kx+6=0kx^2 - 2kx + 6 = 0.
  2. Identify coefficients: a=k,b=−2k,c=6a = k, b = -2k, c = 6.
  3. For equal roots, the discriminant must be zero: D=b2−4ac=0D = b^2 - 4ac = 0.
  4. Substitute: (−2k)2−4(k)(6)=0(-2k)^2 - 4(k)(6) = 0.
  5. Simplify: 4k2−24k=04k^2 - 24k = 0.
  6. Factor: 4k(k−6)=04k(k - 6) = 0.
  7. Possible values for kk are k=0k = 0 or k=6k = 6.
  8. Since a≠0a \neq 0 in a quadratic equation, kk cannot be 00. Therefore, k=6k = 6.

Explanation:

Equal roots imply D=0D = 0. We solve the resulting quadratic equation in terms of kk and exclude values that would make the original equation non-quadratic (where the x2x^2 coefficient would be zero).