f(x) = 4x4 – 21x2 + 27, then the zeroes of f(x) are
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4x^4 + 21x^2 + 27
Let x^2 = a
= 4a^2 - 21a + 27
= 4a^2 - 12a - 9a + 27
= 4a (a - 3) - 9 ( a - 3)
= ( 4a - 9)(a-3)
= ( 4x^2 - 9) (x^2 - 3)
= (2x + 3)(2x-3) (x^2 - 3)
Zeroes are 3/2, - 3/2, + root3 and - root3
Let x^2 = a
= 4a^2 - 21a + 27
= 4a^2 - 12a - 9a + 27
= 4a (a - 3) - 9 ( a - 3)
= ( 4a - 9)(a-3)
= ( 4x^2 - 9) (x^2 - 3)
= (2x + 3)(2x-3) (x^2 - 3)
Zeroes are 3/2, - 3/2, + root3 and - root3
ayushsharma7049:
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Answered by
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Solution :-
4x⁴ - 21x² + 27 = 0
⇒ 4x⁴ - 12x² - 9x² + 27 = 0
⇒ 4x²(x² - 3) - 9(x² - 3) = 0
⇒ (4x² - 9) (x² - 3) = 0
⇒ (2x + 3) (2x - 3) (x² - 3) = 0
⇒ 2x + 3 = 0 2x - 3 = 0 x² - 3 = 0
⇒ 2x = - 3 2x = 3 x² = 3
⇒ x = - 3/2 x = 3/2 x = +√3 or -√3
Answer.
4x⁴ - 21x² + 27 = 0
⇒ 4x⁴ - 12x² - 9x² + 27 = 0
⇒ 4x²(x² - 3) - 9(x² - 3) = 0
⇒ (4x² - 9) (x² - 3) = 0
⇒ (2x + 3) (2x - 3) (x² - 3) = 0
⇒ 2x + 3 = 0 2x - 3 = 0 x² - 3 = 0
⇒ 2x = - 3 2x = 3 x² = 3
⇒ x = - 3/2 x = 3/2 x = +√3 or -√3
Answer.
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