Find the zeros of 2✓3x^2-x-✓3
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Answer:
3√2 x² + 13 x + 6√2
= 3√2 x² + 9x + 4x + 6√2
= 3x(√2 x + 3) + 2√2(√2x + 3)
= (√2x + 3)(3x + 2√2)
Zeroes of the polynomial:
(3x + 2√2) = 0
⇒ x = 2√2/3
(√2x + 3) = 0
⇒ x = -3/√2
Sum of the roots = 2√2/3 -3/√2 = -13/3√2
Product of the roots = 2√2/3 × 3/√2 = 2
Sum of the roots = -b/a = -13/3√2
Product of the roots = c/a = 6√2 / 3√2 = 2.
Step-by-step explanation:
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