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Complex Numbers and Quadratic Equations - Quadratic Equations with real coefficients

Grade 11CBSE

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

🔑Concepts

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A quadratic equation with real coefficients is expressed in the standard form ax2+bx+c=0ax^2 + bx + c = 0, where a,b,c∈Ra, b, c \in \mathbb{R} and a≠0a \neq 0. Visually, this equation represents a parabola that opens upwards if a>0a > 0 and downwards if a<0a < 0.

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The Discriminant, denoted as D=b2−4acD = b^2 - 4ac, determines the nature of the roots. When D<0D < 0, the parabola does not intersect or touch the x-axis at any point, indicating that the roots are not real but complex numbers.

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For a quadratic equation with real coefficients, if the discriminant DD is negative, the roots are complex and always occur in conjugate pairs. This means if one root is z=p+iqz = p + iq, then the other root must be zˉ=p−iq\bar{z} = p - iq.

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The Fundamental Theorem of Algebra states that a polynomial equation of degree nn has exactly nn roots in the complex number system. For quadratic equations (n=2n=2), there are exactly two roots, which may be real and distinct, real and equal, or complex conjugates.

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When solving ax2+bx+c=0ax^2 + bx + c = 0 with D<0D < 0, we introduce the imaginary unit i=−1i = \sqrt{-1}. The square root of the negative discriminant D\sqrt{D} is expressed as i∣D∣i\sqrt{|D|} or i4ac−b2i\sqrt{4ac - b^2}, allowing us to find solutions in the complex plane.

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Graphically, the complex roots of ax2+bx+c=0ax^2 + bx + c = 0 can be visualized as points in the Argand plane. Because they are conjugates (p±iqp \pm iq), they are reflections of each other across the real (horizontal) axis.

📐Formulae

Standard Form: 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}

Roots when D<0D < 0: x=−b±i4ac−b22ax = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}

Sum of roots (α+β\alpha + \beta): −ba-\frac{b}{a}

Product of roots (αβ\alpha \beta): ca\frac{c}{a}

💡Examples

Problem 1:

Solve the quadratic equation x2+3x+5=0x^2 + 3x + 5 = 0 over the set of complex numbers.

Solution:

Step 1: Identify coefficients: a=1,b=3,c=5a = 1, b = 3, c = 5. \nStep 2: Calculate the discriminant D=b2−4ac=32−4(1)(5)=9−20=−11D = b^2 - 4ac = 3^2 - 4(1)(5) = 9 - 20 = -11. \nStep 3: Since D<0D < 0, the roots are complex. Use the formula x=−b±i∣D∣2ax = \frac{-b \pm i\sqrt{|D|}}{2a}. \nStep 4: Substitute values: x=−3±i112(1)x = \frac{-3 \pm i\sqrt{11}}{2(1)}. \nStep 5: The roots are x=−3+i112x = \frac{-3 + i\sqrt{11}}{2} and x=−3−i112x = \frac{-3 - i\sqrt{11}}{2}.

Explanation:

We use the quadratic formula for case where the discriminant is negative. The negative sign under the square root is replaced by the imaginary unit ii outside the radical.

Problem 2:

Solve 2x2+x+2=0\sqrt{2}x^2 + x + \sqrt{2} = 0.

Solution:

Step 1: Identify coefficients: a=2,b=1,c=2a = \sqrt{2}, b = 1, c = \sqrt{2}. \nStep 2: Calculate D=b2−4ac=12−4(2)(2)=1−4(2)=1−8=−7D = b^2 - 4ac = 1^2 - 4(\sqrt{2})(\sqrt{2}) = 1 - 4(2) = 1 - 8 = -7. \nStep 3: Apply the quadratic formula for complex roots: x=−1±−722x = \frac{-1 \pm \sqrt{-7}}{2\sqrt{2}}. \nStep 4: Simplify using ii: x=−1±i722x = \frac{-1 \pm i\sqrt{7}}{2\sqrt{2}}.

Explanation:

Even with irrational coefficients, the process remains the same. Since D=−7D = -7 is negative, the roots are a pair of complex conjugates.