⇔ 16x⁴ − 49x² + 3 = 0
⇔ 16y² − 49y + 3 = 0 ∧ y = x²
⇔ (y − 3)(16y − 1) = 0 ∧ y = x²
⇔ [y = 3 ∨ y =
] ∧ y = x² ⇔ x² = 3 ∨ x² =
⇔ x = ±
∨ x = ± 
twee irrationale en twee rationale getallen
|
De oplossingenverzameling van de bikwadratische vergelijking 16x4 = 49x2 − 3 bestaat uit |
A. 2 rationale getallen |
|---|---|
| B. 2 rationale en 2 irrationale getallen | |
| C. 2 irrationale getallen | |
| D. 2 rationale en 1 irrationaal getal | |
| E. 2 irrationale en 1 rationaal getal |
[ 5-4170 - op net sinds 15.12.06-(E)-20.7.2024 ]
|
The set of solutions of the biquadratic equation 16x4 = 49x2 − 3 consists of |
A. 2 rational numbers |
|---|---|
| B. 2 rational numbers and 2 irrational numbers | |
| C. 2 irrational numbers | |
| D. 2 rational numbers and 1 irrational number | |
| E. 2 irrational numbers and 1 rational number |
] ∧ y = x² ∨ x = ± 