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Algebra - Solving quadratic equations by factoring and using the quadratic formula

Grade 10IB

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

🔑Concepts

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A quadratic equation is an equation of the form ax2+bx+c=0ax^2 + bx + c = 0, where a≠0a \neq 0. Visually, the graph of a quadratic equation is a U-shaped curve called a parabola. If a>0a > 0, the parabola opens upwards like a cup; if a<0a < 0, it opens downwards like a cap.

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The Zero Product Property states that if the product of two factors is zero, then at least one of the factors must be zero. For example, if (x−p)(x−q)=0(x - p)(x - q) = 0, then x−p=0x - p = 0 or x−q=0x - q = 0. Visually, these values pp and qq represent the x-intercepts where the parabola crosses the horizontal x-axis.

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Factoring Monic Trinomials involves solving x2+bx+c=0x^2 + bx + c = 0 by finding two numbers that multiply to give cc and add to give bb. Once found, the equation is written as (x+m)(x+n)=0(x + m)(x + n) = 0, allowing us to solve for xx.

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Factoring Non-Monic Trinomials (where a≠1a \neq 1) often requires 'splitting the middle term.' We look for two numbers that multiply to a×ca \times c and add to bb. This allows the equation to be grouped and factored into two binomials.

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The Quadratic Formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} is a universal method used to find the roots of any quadratic equation, especially when the expression cannot be easily factored into integers.

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The Discriminant, denoted as Δ=b2−4ac\Delta = b^2 - 4ac, determines the nature of the roots. Visually: if Δ>0\Delta > 0, the parabola intersects the x-axis at two distinct points; if Δ=0\Delta = 0, the vertex of the parabola touches the x-axis at exactly one point; if Δ<0\Delta < 0, the parabola never touches the x-axis, resulting in no real solutions.

📐Formulae

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

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

Discriminant: Δ=b2−4ac\Delta = b^2 - 4ac

Difference of Two Squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)

💡Examples

Problem 1:

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

Solution:

Step 1: Identify two numbers that multiply to 66 and add to −5-5. These numbers are −2-2 and −3-3. Step 2: Rewrite the equation in factored form: (x−2)(x−3)=0(x - 2)(x - 3) = 0. Step 3: Apply the Zero Product Property: x−2=0x - 2 = 0 or x−3=0x - 3 = 0. Step 4: Solve for xx: x=2x = 2 or x=3x = 3.

Explanation:

This method works because the leading coefficient is 1. We look for factors of the constant term that sum to the middle coefficient.

Problem 2:

Solve 2x2+4x−1=02x^2 + 4x - 1 = 0 using the quadratic formula. Round to two decimal places.

Solution:

Step 1: Identify coefficients a=2,b=4,c=−1a = 2, b = 4, c = -1. Step 2: Substitute into the formula: x=−4±42−4(2)(−1)2(2)x = \frac{-4 \pm \sqrt{4^2 - 4(2)(-1)}}{2(2)}. Step 3: Simplify the discriminant: x=−4±16+84=−4±244x = \frac{-4 \pm \sqrt{16 + 8}}{4} = \frac{-4 \pm \sqrt{24}}{4}. Step 4: Calculate numerical values: x=−4±4.8994x = \frac{-4 \pm 4.899}{4}. Step 5: Solve for both cases: x1≈0.8994≈0.22x_1 \approx \frac{0.899}{4} \approx 0.22 and x2≈−8.8994≈−2.22x_2 \approx \frac{-8.899}{4} \approx -2.22.

Explanation:

The quadratic formula is used here because the equation does not have simple integer factors. The ±\pm symbol ensures we find both possible x-intercepts of the parabola.