Asked by Bad "All-American" Bunny
show that fundamental theorem of algebra is true for 2x^2-12x+18=0 how many rational non rational and unique solutions
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GPT-5 mini
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Start with the quadratic
2x^2 − 12x + 18 = 0.
Divide by 2: x^2 − 6x + 9 = 0, which factors as (x − 3)^2 = 0. So x = 3 is a root of multiplicity 2.
Check with the discriminant: Δ = b^2 − 4ac = (−12)^2 − 4·2·18 = 144 − 144 = 0, so there is one repeated real root.
By the Fundamental Theorem of Algebra a degree-2 polynomial has exactly two complex roots counted with multiplicity. Here both roots are 3 (a real number), so:
- Total complex roots (counting multiplicity): 2 (both x = 3).
- Distinct real roots: 1 (x = 3).
- Rational roots: 1 distinct rational root (x = 3); counting multiplicity there are 2 rational roots.
- Non‑rational (irrational) roots: 0.
2x^2 − 12x + 18 = 0.
Divide by 2: x^2 − 6x + 9 = 0, which factors as (x − 3)^2 = 0. So x = 3 is a root of multiplicity 2.
Check with the discriminant: Δ = b^2 − 4ac = (−12)^2 − 4·2·18 = 144 − 144 = 0, so there is one repeated real root.
By the Fundamental Theorem of Algebra a degree-2 polynomial has exactly two complex roots counted with multiplicity. Here both roots are 3 (a real number), so:
- Total complex roots (counting multiplicity): 2 (both x = 3).
- Distinct real roots: 1 (x = 3).
- Rational roots: 1 distinct rational root (x = 3); counting multiplicity there are 2 rational roots.
- Non‑rational (irrational) roots: 0.
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