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Quadratic Equations - Solution of Quadratic Equations by Quadratic Formula

Grade 10CBSE

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

🔑Concepts

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A quadratic equation in the variable xx is an equation of the form ax2+bx+c=0ax^2 + bx + c = 0, where a,b,ca, b, c are real numbers and a≠0a \neq 0. Graphically, this represents a parabola, a symmetrical 'U-shaped' curve that opens upwards if a>0a > 0 and downwards if a<0a < 0.

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The roots of a quadratic equation are the values of xx that make the equation true. On a graph, these roots correspond to the x-coordinates of the points where the parabola intersects the x-axis.

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The expression D=b2−4acD = b^2 - 4ac is known as the Discriminant. It is a crucial tool used to determine the nature of the roots without solving the entire equation.

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If the discriminant D>0D > 0, the equation has two distinct real roots. Visually, the parabola crosses the x-axis at two different points.

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If the discriminant D=0D = 0, the equation has two equal real roots (often called a single repeated root). In this case, the parabola's vertex touches the x-axis at exactly one point.

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If the discriminant D<0D < 0, the equation has no real roots. This means the parabola lies entirely above or entirely below the x-axis and never touches it.

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The Quadratic Formula is an algebraic method derived from the process of 'completing the square.' It is used to find the roots of any quadratic equation, especially when the equation cannot be easily factored by splitting the middle term.

📐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=−b2ax = \frac{-b}{2a}

💡Examples

Problem 1:

Solve the quadratic equation x2−5x+6=0x^2 - 5x + 6 = 0 using the quadratic formula.

Solution:

Step 1: Identify coefficients a,b,a, b, and cc. Here, a=1a = 1, b=−5b = -5, and c=6c = 6. \nStep 2: Calculate the discriminant DD. \nD=b2−4acD = b^2 - 4ac \nD=(−5)2−4(1)(6)D = (-5)^2 - 4(1)(6) \nD=25−24=1D = 25 - 24 = 1. \nStep 3: Since D>0D > 0, the equation has two distinct real roots. \nStep 4: Apply the quadratic formula: \nx=−(−5)±12(1)x = \frac{-(-5) \pm \sqrt{1}}{2(1)} \nx=5±12x = \frac{5 \pm 1}{2} \nStep 5: Find the two values of xx. \nx1=5+12=62=3x_1 = \frac{5 + 1}{2} = \frac{6}{2} = 3 \nx2=5−12=42=2x_2 = \frac{5 - 1}{2} = \frac{4}{2} = 2 \nRoots are x=3x = 3 and x=2x = 2.

Explanation:

We first extracted the coefficients from the standard form equation, calculated the discriminant to confirm the nature of the roots, and then substituted the values into the quadratic formula to find the two specific solutions.

Problem 2:

Find the roots of the equation 3x2−2x−1=03x^2 - 2x - 1 = 0.

Solution:

Step 1: Identify coefficients: a=3,b=−2,c=−1a = 3, b = -2, c = -1. \nStep 2: Calculate D=b2−4acD = b^2 - 4ac. \nD=(−2)2−4(3)(−1)D = (-2)^2 - 4(3)(-1) \nD=4+12=16D = 4 + 12 = 16. \nStep 3: Apply the quadratic formula: \nx=−(−2)±162(3)x = \frac{-(-2) \pm \sqrt{16}}{2(3)} \nx=2±46x = \frac{2 \pm 4}{6} \nStep 4: Solve for both cases: \nx=2+46=66=1x = \frac{2 + 4}{6} = \frac{6}{6} = 1 \nx=2−46=−26=−13x = \frac{2 - 4}{6} = \frac{-2}{6} = -\frac{1}{3} \nRoots are x=1x = 1 and x=−13x = -\frac{1}{3}.

Explanation:

The discriminant was a perfect square (16), which indicates that the roots are rational. Using the quadratic formula allows for a straightforward calculation even when the leading coefficient (aa) is greater than 1.