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Artist 52 [7]
3 years ago
5

If (2-square root of 3 ) is a root of a polynomial with integer coefficients which of the follow must be another root

Mathematics
1 answer:
Anika [276]3 years ago
4 0

\text{If k and l are the roots of a polynomial w(x), then}\ w(x)=a(x-k)(x-l).

(2-\sqrt3)-a\ root\ of\ a\ polynomial\\\\another\ root\ must\ be\ C.\ (2+\sqrt3),\ because\\\\\ [x-(2-\sqrt3)][x-(2+\sqrt3)]=(x-2+\sqrt3)(x-2-\sqrt3)\\\\=[(x-2)+\sqrt3][(x-2)-\sqrt3]\\\\\text{use}\ (a-b)(a+b)=a^2-b^2\\\\=(x-2)^2-(\sqrt3)^2\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=x^2-2(x)(2)+2^2-3=x^2-4x+4-3=x^2-4x+1

Answer:\ \boxed{C.\ (2+\sqrt3)}

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Solve the compound inequality |3x-9|≤15 and |2x-3|≥5. Give answer in interval notation.
laila [671]

Answer:

The solution of |3x-9|≤15 is [-2;8] and the solution |2x-3|≥5 of is  (-∞,2] ∪ [8,∞)

Step-by-step explanation:

When solving absolute value inequalities, there are two cases to consider.

Case 1: The expression within the absolute value symbols is positive.

Case 2: The expression within the absolute value symbols is negative.

The solution is the intersection of the solutions of these two cases.

In other words, for any real numbers a and b,

  • if |a|> b then a>b or a<-b
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So, being |3x-9|≤15

Solving: 3x-9 ≤ 15

3x ≤15 + 9

3x ≤24

x ≤24÷3

x≤8

or 3x-9 ≥ -15

3x ≥-15 +9

3x ≥-6

x ≥ (-6)÷3

x ≥ -2

The solution is made up of all the intervals that make the inequality true. Expressing the solution as an interval: [-2;8]

So, being |2x-3|≥5

Solving: 2x-3 ≥ 5

2x ≥ 5 + 3

2x ≥8

x ≥8÷2

x≥8

or 2x-3 ≤ -5

2x ≤-5 +3

2x ≤-2

x ≤ (-2)÷2

x ≤ -2

Expressing the solution as an interval: (-∞,2] ∪ [8,∞)

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x = 13.127 cm

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Answer:

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m_a_m_a [10]

Answer:

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