Quadratic formula, Discriminant, and Vieta's formulas

LVL: FREE

MODULE: Polynomials and Functions

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✖️ 1. Deriving and applying the Quadratic Formula to find exact roots

🔑 The Quadratic Formula

  • For any quadratic ax2+bx+c=0ax^2 + bx + c = 0 where a0a \neq 0, the roots are x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
  • The plus-minus symbol ±\pm gives you two solutions in one formula.
  • Always identify aa, bb, and cc first before plugging into the formula.
  • The formula works even when factoring is impossible or messy.
  • If aa is negative, factor it out or keep careful track of signs.

Example: Solve 2x25x+2=02x^2 - 5x + 2 = 0. Here a=2a=2, b=5b=-5, c=2c=2. Then x=5±25164=5±34x = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}, so x=2x = 2 or x=0.5x = 0.5.

💡 Memory hook: "Negative b, plus or minus the square root, all over 2a."

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1. Deriving and applying the Quadratic Formula to find exact roots

Deriving and Applying the Quadratic Formula

The quadratic formula x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} provides the exact roots of any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 where a0a \neq 0. It is derived by completing the square on the general form.

Intuition: The formula encodes all possible solutions by accounting for the parabola's vertex position and width, yielding two roots (possibly equal or complex) through the ±\pm operation.

Core Rules:

  • Applicability: Valid only when a0a \neq 0 (otherwise the equation is linear).
  • Two solutions: The ±\pm symbol generates both roots simultaneously.
  • Exact form: Roots may be irrational or complex; leave in radical form unless approximation is requested.
  • Sign of aa: Does not affect root existence, only parabola orientation.

Consequence: Every quadratic equation has exactly two roots in the complex number system, though they may coincide or be non-real.

Example: For 2x24x6=02x^2 - 4x - 6 = 0, we have x=4±16+484=4±84x = \frac{4 \pm \sqrt{16 + 48}}{4} = \frac{4 \pm 8}{4}, giving x=3x = 3 or x=1x = -1.

TASK_1[0 / 3]
LVL_2
RSN: DEBUG

A student is solving the equation x25x6=0x^2 - 5x - 6 = 0 using the quadratic formula. They write their first step as:

x=(5±2524)/2x = (5 \pm \sqrt{25 - 24}) / 2

Based on the core rules of the formula, what is their specific error?

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✖️ 2. Using the Discriminant to determine the number and nature of roots

🔍 The Discriminant Test

  • The discriminant is D=b24acD = b^2 - 4ac (the part under the square root).
  • If D>0D > 0, you get two distinct real roots.
  • If D=0D = 0, you get exactly one real root (a repeated root).
  • If D<0D < 0, you get no real roots (two complex roots).
  • You can check the discriminant before solving to know what to expect.

Example: For x2+4x+5=0x^2 + 4x + 5 = 0, compute D=1620=4<0D = 16 - 20 = -4 < 0, so no real solutions exist.

💡 Visual cue: Positive D = parabola crosses x-axis twice; Zero D = touches once; Negative D = floats above or below.

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2. Using the Discriminant to determine the number and nature of roots

Using the Discriminant (D=b24acD = b^2 - 4ac)

The discriminant D=b24acD = b^2 - 4ac is the expression under the square root in the quadratic formula. It determines the number and type of roots without computing them explicitly.

Intuition: The discriminant measures whether the parabola intersects the xx-axis (real roots), touches it (repeated root), or misses it entirely (complex roots).

Core Rules:

  • D>0D > 0: Two distinct real roots.
  • D=0D = 0: Exactly one repeated real root (the vertex touches the xx-axis).
  • D<0D < 0: Two complex conjugate roots (no real intersections).
  • Perfect square DD: If DD is a perfect square, roots are rational; otherwise irrational.

Consequence: The discriminant provides immediate qualitative information about solutions, essential for analyzing feasibility in applied problems.

Example: For x2+2x+5=0x^2 + 2x + 5 = 0, we have D=420=16<0D = 4 - 20 = -16 < 0, so the equation has two complex roots and no real solutions.

TASK_1[0 / 3]
LVL_2
RSN: LOGIC

Calculate the discriminant of the quadratic equation 3x22x5=03x^2 - 2x - 5 = 0.

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✖️ 3. Using Vieta's formulas to find sums and products of roots without solving

🧮 Vieta's Shortcut Formulas

  • For ax2+bx+c=0ax^2 + bx + c = 0 with roots rr and ss, sum of roots is r+s=bar + s = -\frac{b}{a}.
  • The product of roots is rs=cars = \frac{c}{a}.
  • These formulas let you find relationships without computing the actual roots.
  • Useful for checking answers or solving problems about root properties.
  • Convention: Write the equation in standard form first so you identify aa, bb, cc correctly.

Example: For 3x212x+9=03x^2 - 12x + 9 = 0, sum =123=4= \frac{12}{3} = 4 and product =93=3= \frac{9}{3} = 3.

💡 Memory hook: Sum uses b-b, product uses cc, both divided by aa.

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3. Using Vieta's formulas to find sums and products of roots without solving

Using Vieta's Formulas

Vieta's formulas relate the coefficients of ax2+bx+c=0ax^2 + bx + c = 0 to its roots r1r_1 and r2r_2 without solving: r1+r2=bar_1 + r_2 = -\frac{b}{a} and r1r2=car_1 \cdot r_2 = \frac{c}{a}. These follow from expanding (xr1)(xr2)=0(x - r_1)(x - r_2) = 0.

Intuition: The sum of roots depends on the linear coefficient, while the product depends on the constant term, both scaled by the leading coefficient.

Core Rules:

  • Sum: r1+r2=bar_1 + r_2 = -\frac{b}{a} (note the negative sign).
  • Product: r1r2=car_1 \cdot r_2 = \frac{c}{a}.
  • Validity: Holds for all roots, real or complex.
  • Sign interpretation: If ca>0\frac{c}{a} > 0, roots have the same sign; if ca<0\frac{c}{a} < 0, opposite signs.

Consequence: Vieta's formulas enable rapid analysis of root properties and verification of solutions without explicit computation.

Example: For 3x212x+9=03x^2 - 12x + 9 = 0, the sum is 123=4-\frac{-12}{3} = 4 and product is 93=3\frac{9}{3} = 3, so roots are r1=3,r2=1r_1 = 3, r_2 = 1.

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STRC: REVERSE

For the equation 2x210x+7=02x^2 - 10x + 7 = 0, what is the sum of its roots?

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✖️ 4. Reconstructing a quadratic equation given its roots

🔨 Building Quadratics from Roots

  • If roots are rr and ss, the quadratic is (xr)(xs)=0(x - r)(x - s) = 0.
  • Expand to get x2(r+s)x+rs=0x^2 - (r+s)x + rs = 0 using Vieta's formulas in reverse.
  • You can multiply the entire equation by any nonzero constant and it remains valid.
  • This method works for any two numbers, real or complex.
  • Always expand and simplify to standard form.

Example: Roots are 2 and -3. Then (x2)(x+3)=x2+x6=0(x - 2)(x + 3) = x^2 + x - 6 = 0.

💡 Quick check: Plug each root back into your equation to verify it equals zero.

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4. Reconstructing a quadratic equation given its roots

Reconstructing a Quadratic Equation from Roots

Given roots r1r_1 and r2r_2, the quadratic equation is (xr1)(xr2)=0(x - r_1)(x - r_2) = 0, which expands to x2(r1+r2)x+r1r2=0x^2 - (r_1 + r_2)x + r_1 r_2 = 0. This is the reverse application of Vieta's formulas.

Intuition: A quadratic is uniquely determined (up to scaling) by its two roots, as the factored form directly encodes the xx-intercepts.

Core Rules:

  • Factored form: Start with (xr1)(xr2)=0(x - r_1)(x - r_2) = 0.
  • Expanded form: x2Sx+P=0x^2 - Sx + P = 0 where S=r1+r2S = r_1 + r_2 and P=r1r2P = r_1 r_2.
  • Scaling: Multiply by any nonzero constant aa to obtain ax2aSx+aP=0ax^2 - aSx + aP = 0.
  • Irrational/complex roots: The method works identically; coefficients may be irrational or complex.

Consequence: This technique is essential for constructing equations from geometric or physical constraints.

Example: Given roots 55 and 2-2, we have S=3S = 3, P=10P = -10, yielding x23x10=0x^2 - 3x - 10 = 0.

TASK_1[0 / 3]
LVL_2
STRC: REVERSE

A quadratic equation has roots 44 and 77. Which of the following represents this equation in the expanded form x2Sx+P=0x^2 - Sx + P = 0?

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✖️ 5. Applications: Analyzing stability and optimization with quadratics

🌍 Real-World Quadratic Applications

  • Population models: Quadratics model growth with limiting factors; roots show equilibrium points.
  • Pricing problems: Revenue R=(price)(quantity)R = (price)(quantity) is often quadratic; maximum occurs at the vertex.
  • Use the discriminant to check if a model has real solutions (feasible scenarios).
  • Use Vieta's formulas to quickly find total or average outcomes without full solving.
  • The vertex formula x=b2ax = -\frac{b}{2a} finds optimal price or population level.

Example: Revenue R=2p2+40pR = -2p^2 + 40p maximizes at p=404=10p = \frac{40}{4} = 10 dollars per unit.

💡 Context cue: Negative aa means parabola opens down, so vertex is a maximum.

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5. Applications: Analyzing stability and optimization with quadratics

Applications: Stability and Optimization

Quadratic equations model equilibrium points in population dynamics (e.g., logistic growth) and revenue functions in economics. The discriminant determines feasibility, while Vieta's formulas reveal stability conditions.

Intuition: Real roots correspond to physically meaningful equilibria or break-even points; the discriminant indicates whether such states exist.

Core Rules:

  • Population models: Roots of rN(1N/K)=hrN(1 - N/K) = h (harvest rate hh) determine sustainable populations; D<0D < 0 implies extinction.
  • Revenue optimization: For R(p)=ap2+bpR(p) = -ap^2 + bp, the vertex p=b2ap = \frac{b}{2a} maximizes revenue; roots show break-even prices.
  • Stability: In discrete models, if the product of roots (via Vieta) exceeds 1 in absolute value, equilibria are unstable.
  • Constraint analysis: D=0D = 0 marks critical thresholds (e.g., maximum sustainable harvest).

Consequence: Quadratic analysis provides quantitative predictions for system behavior and optimal decision-making.

Example: For revenue R(p)=2p2+40pR(p) = -2p^2 + 40p, roots are p=0p = 0 and p=20p = 20 dollars; maximum revenue occurs at p=10p = 10 dollars.

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MOD: TRANSLATE

A company's revenue function is given by R(p)=3p2+60pR(p) = -3p^2 + 60p, where pp is the price in dollars. Find the price pp that maximizes the revenue.

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