As per Descartes Theorem, a polynomial of degree n (n is the highest exponent of a polynomial) has n roots (or number of zeros) be it positive, negative, real or complex.
in f(x) = x⁶ + x⁵ + x⁴ + 4x³ − 12x² + 12, the highest degree is 6, then it has
a total of 6 zeros, positive, negative, real or complex.
Terminated. Any decimal that repeats goes on forever. Such as pie: 3.1415...
Answer:
x = 4 or x = (-3)/2
Step-by-step explanation:
Solve for x over the real numbers:
2 x^2 - 5 x - 12 = 0
x = (5 ± sqrt((-5)^2 - 4×2 (-12)))/(2×2) = (5 ± sqrt(25 + 96))/4 = (5 ± sqrt(121))/4:
x = (5 + sqrt(121))/4 or x = (5 - sqrt(121))/4
sqrt(121) = sqrt(11^2) = 11:
x = (5 + 11)/4 or x = (5 - 11)/4
(5 + 11)/4 = 16/4 = 4:
x = 4 or x = (5 - 11)/4
(5 - 11)/4 = -6/4 = -3/2:
Answer: x = 4 or x = (-3)/2
The number which could be added to both sides of the equation to complete the square is; -2.25.
<h3>Which number could be added to both sides of this quadratic equation to complete the square?</h3>
The quadratic equations give in the task content is; 1=x²-3x. Hence, to express the equation by completing the square method; we have;
1 - 2.25 = (x -3/2)² - 2.25
Hence, the number which should be added to both sides of the equation is; -2.25.
Read more on completing the square;
brainly.com/question/10449635
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