Verified answer

Hello again,

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0.► In any quadratic equation of the form:

   𝑥² – 𝑆𝑥 + 𝑃 = 0

we always have the properties:

   ● 𝑆 will be the sum of the roots.

   ● 𝑃 will be the product of the roots.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

2.► Thus starting from your equation:

   3𝑥² + 𝑞𝑥 – 12 = 0

   𝑥² + (𝑞/3)𝑥 – 4 = 0

So let 𝛼 be one of the root of this equation. The other, differing by 4, will then be: 𝛼+4.

We then have:

   { 𝑆 = 2𝛼 + 4 = -𝑞/3

   { 𝑃 = 𝛼² + 4𝛼 = -4

From last equation:

    𝛼² + 4𝛼 = -4

    𝛼² + 4𝛼 + 4 = 0

    (𝛼 + 2)² = 0

    𝛼 + 2 = 0

    𝛼 = -2

    𝛼 = -2

From first equation:

   2𝛼 + 4 = -𝑞/3

   𝑞/3 = -2𝛼 – 4

   𝑞 = -6𝛼 – 12

and knowing 𝛼=-2

   𝑞 = 0     ◄◄◄ ANSWER

Regards,

Dragon.Jade :-)

3x² - qx + 12 = 0

Polynomial like: ax² + bx + c, where:

a = 3

b = - q

c = 12

Δ = b² - 4ac (discriminant)

x₁ = (- b - √Δ)/2a

x₂ = (- b + √Δ)/2a

x₁ - x₂ = 4

[(- b - √Δ)/2a] - [(- b + √Δ)/2a] = 4

[(- b - √Δ) - (- b + √Δ)]/2a = 4

(- b - √Δ) - (- b + √Δ) = 8a

- b - √Δ + b - √Δ = 8a

- 2√Δ = 8a

√Δ = - 4a

(√Δ)² = (- 4a)²

Δ = 16a² → recall: Δ = b² - 4ac

b² - 4ac = 16a² → recall: a = 3

b² - 12c = 144 → recall: c = 12

b² - 144 = 144

b² = 288 → recall: b = q

q² = 288

q = ± √288

q = ± 12√2

3x²+qx-12=0

r₁ = [-q + √(q² + 4•3•12)}/6 = [-q + √(q²+144)]/6

r₂ = [-q - √(q² + 4•3•12)}/6 = [-q -√(q²+144)]/6

Given r₁ - r₂ = 4

r₁ - r₂ = [-q + √(q²+144)]/6 - [-q -√(q²+144)]/6

= 2√(q²+144)/6 = 4

√(q²+144) = 12

q² + 144 = 144

q² = 0

Ans: q = 0

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Verification:

3x² + (0)x - 12 = 0

x² = 4

r₁ = 2

r₂ = -2

r₁ - r₂ = 4

q could be any number

do you mean find p?