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Ugo [173]
1 year ago
7

The steps for inscribing a circle in a triangle are shown:Step 1Construct the perpendicular bisector of one angle of the triangl

e.Step 2Construct the angle bisector of another angle of the triangle.Step 3Indicate the point of intersection of the bisectors with a point representing the center of the circle.Step 4Construct a perpendicular line from the center to a side of the triangle.Step 5Place the compass on the center, adjust its length to where the perpendicular line intersects a side of the triangle, and draw your inscribed circle.Which step is incorrect, and how can it be fixed?
Mathematics
1 answer:
nevsk [136]1 year ago
6 0

Solution

- The correct steps for drawing a circle in a triangle are given below:

1. Bisect one of the angles

2. Bisect another angle

3. Where they cross is the center of the inscribed circle, called the incenter

4. Construct a perpendicular from the center point to one side of the triangle

5. Place compass on the center point, adjust its length to where the perpendicular crosses the triangle, and draw your inscribed circle!

- Based on these steps above and comparing them with the steps given in the question, every step given in the question is correct except Step 1.

- Step 1 given in the question, says we should construct a perpendicular bisector of one angle in the triangle. This is not correct because of the use of perpendicular bisector. This is only used for bisecting lines, not angles.

Final Answer

The answer is Step 1

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Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.
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A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
  • The maximum number of roots of a polynomial of degree n is n.
  • For a polynomial with real coefficients, the roots can be real or complex.
  • The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if a+ib is a root, then a-ib is also a root.

If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

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