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Serhud [2]
3 years ago
11

Find the value of k such that the quadratic polynomials x2-(k+6)x+2(2k+1) as sum of the zeroes as the half of their product

Mathematics
1 answer:
Gwar [14]3 years ago
6 0

Answer:

  k = 5

Step-by-step explanation:

The sum of the zeros is the opposite of the coefficient of x, so is (k+6).

The product of zeros is the constant term, 2(2k+1), so half their product is (2k+1).

The problem statement asks us to find k so that these values are the same:

  k +6 = 2k +1

  5 = k . . . . . . . . subtract k+1

The value of k is 5.

_____

The zeros are 5.5±√8.25. Their sum is 11; their product is 22.

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In Triangle BCD, BC=BD, and angle B =13x-35, angle C = 5x-19, angle D=2x+14. Find the measure of each angle
aleksandr82 [10.1K]

Answer:

B = 108

C = 36

D = 36

Step-by-step explanation:

Given

B =13x - 35

C = 5x - 19

D=2x+14

BC = BD

Required

Determine the measure of each angle

The angles in a triangle sum up t 180. i.e.

B + C + D = 180

Since BC = BD, then C = D

B + C + D = 180 can then be rewritten as:

B + C + C = 180

B + 2C = 180

Substitute values for B and C

13x - 35 + 2(5x - 19) = 180

Open Bracket

13x - 35 + 10x - 38 = 180

Collect Like Terms

13x + 10x = 180 + 38 + 35

23x= 253

Divide both sides by 23

x = 11

To get the measure of each angle, substitute 11 for x in the expressions of B, C and D

B =13x - 35

B = 13 * 11 - 35

B = 108

C = 5x - 19

C = 5 * 11 - 19

C = 36

D=2x+14

D = 2 * 11 + 14

D = 36

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