Consider the quadratic equation x2 = 4x - 5. How many solutions does the equation have?

Step-by-step explanation:

The equation can be rearranged to vertex form:

  x^2 -4x = -5 . . . . . . . . . subtract 4x

  x^2 -4x +4 = -5 +4 . . . . add 4

  (x -2)^2 = -1 . . . . . . . . . show the left side as a square

  x -2 = ±√-1 = ±i . . . . . . take the square root; the right side is imaginary

  x = 2 ± i . . . . . . . . . . . . . add 2. These are the complex solutions.

Comment on the question

Every 2nd degree polynomial equation has two solutions. They may be real, complex, or (real and) identical. That is, there may be 0, 1, or 2 real solutions. This equation has 0 real solutions, because they are both complex.

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step-by-step explanation:

geometry: triangles

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answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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