✨ Math Magic: Polynomial Adventures ✨

1️⃣ Nature of Roots

If k is real, discuss the nature of the roots of the polynomial equation:

(2x² + kx + k)/2 = 0

Step 1: Simplify the Equation

First, let's simplify the equation:

(2x² + kx + k)/2 = 0

Multiply both sides by 2:

2x² + kx + k = 0

Step 2: Understand the Quadratic

This is a quadratic equation of the form:

ax² + bx + c = 0

Where:

a = 2, b = k, c = k

Step 3: Find the Discriminant

The nature of roots is determined by the discriminant (D):

D = b² - 4ac

Substitute our values:

D = k² - 4(2)(k) = k² - 8k

Step 4: Analyze the Discriminant

Let's find when D is positive, zero, or negative:

k² - 8k > 0 → k(k - 8) > 0

This inequality holds when k < 0 or k > 8

k² - 8k = 0 → k = 0 or k = 8

k² - 8k < 0 → 0 < k < 8

✨ Final Answer:

  • When k < 0 or k > 8: Two distinct real roots (D > 0)
  • When k = 0 or k = 8: One real root (D = 0)
  • When 0 < k < 8: No real roots (D < 0)

2️⃣ Polynomial with Complex Root

Find a polynomial equation of minimum degree with rational coefficients, having:

2 + √3i

as a root.

Step 1: Identify the Conjugate

For polynomials with rational coefficients, complex roots come in conjugate pairs.

If 2 + √3i is a root, then 2 - √3i must also be a root.

Step 2: Form the Factors

Create factors from the roots:

x - (2 + √3i)

x - (2 - √3i)

Step 3: Multiply the Factors

Multiply them together:

[x - (2 + √3i)][x - (2 - √3i)]

This is a difference of squares pattern:

(x - 2)² - (√3i)²

✨ Final Answer:

Expand the expression:

(x - 2)² - (√3i)² = x² - 4x + 4 - (3i²)

Since i² = -1:

x² - 4x + 4 - (3)(-1) = x² - 4x + 4 + 3 = x² - 4x + 7

The minimal polynomial is:

x² - 4x + 7 = 0

3️⃣ Another Complex Root

Find a polynomial equation of minimum degree with rational coefficients, having:

3 + 2i

as a root.

Step 1: Identify the Conjugate

The conjugate root is 3 - 2i.

Step 2: Form the Factors

[x - (3 + 2i)][x - (3 - 2i)]

Step 3: Multiply

(x - 3)² - (2i)² = x² - 6x + 9 - (4i²)

= x² - 6x + 9 - (4)(-1) = x² - 6x + 9 + 4 = x² - 6x + 13

✨ Final Answer:

x² - 6x + 13 = 0

4️⃣ Polynomial with Irrational Root

Find a polynomial equation of minimum degree with rational coefficients, having:

√5 - √3

as a root.

Step 1: Set x = √5 - √3

x = √5 - √3

Step 2: Isolate the Radicals

x + √3 = √5

Step 3: Square Both Sides

(x + √3)² = (√5)²

x² + 2√3x + 3 = 5

Step 4: Isolate Remaining Radical

x² + 2√3x = 2

2√3x = 2 - x²

Step 5: Square Again

(2√3x)² = (2 - x²)²

12x² = 4 - 4x² + x⁴

✨ Final Answer:

x⁴ - 16x² + 4 = 0

5️⃣ Line and Parabola Intersection

Prove that a straight line and parabola cannot intersect at more than two points.

Step 1: General Equations

Let's consider:

Parabola: y = ax² + bx + c (a ≠ 0)

Line: y = mx + d

Step 2: Find Intersection Points

Set the y-values equal:

ax² + bx + c = mx + d

Rearrange:

ax² + (b - m)x + (c - d) = 0

Step 3: Nature of Solutions

This is a quadratic equation in x. The maximum number of real solutions is 2.

Therefore, a line and parabola can intersect at:

  • 2 points (when discriminant > 0)
  • 1 point (when discriminant = 0, tangent)
  • 0 points (when discriminant < 0)

✨ Conclusion:

A straight line and parabola can intersect at most at two points.