How do you solve for n^3 + 3n^2 + n - 33 = 0?

  one real root: n ≈ 2.38450287889

Step-by-step explanation:

My favorite solution method for higher-degree polynomials is to use a graphing calculator.

Descartes' rule of signs tells you the one sign change among coefficients means there will be one positive real root. A graph shows you it is about 2.4, hence irrational (not a divisor of 33, so not rational).

You can use a cubic formula to find an explicit expression for the root, or you can find its value using any of several iteration methods. The attachment shows Newton's method iteration being used to refine the graph value of 2.385 to the more accurate 2.38450287889.

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Factoring that root from the cubic results in a quadratic with irrational coefficients. Its vertex form is approximately ...

  y = (n +2.692)² + 6.591

Hence, the complex roots will be near -2.692±i√6.591.

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There are formulas for the roots of a cubic. The formula tells you the real root for this cubic is ...

  n = 2√(2/3)cosh(1/3·arccosh(24√(3/2))) -1 ≈ 2.38450287889

Hey mate where is the graph

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)