ABCD is a quadrilateral. AB=CD Angle ABC = angle BCD Prove that AC = BD
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<h3>Answer:</h3>
(x, y) ≈ (1.49021612010, 1.22074408461)
<h3>Explanation:</h3>This is best solved graphically or by some other machine method. The approximate solution (x=1.49, y=1.221) can be iterated by any of several approaches to refine the values to the ones given above. The values above were obtained using Newton's method iteration.
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Setting the y-values equal and squaring both sides of the equation gives ...
... √x = x² -1
... x = (x² -1)² = x⁴ -2x² +1 . . . . . square both sides
... x⁴ -2x² -x +1 = 0 . . . . . polynomial equation in standard form.
By Descarte's rule of signs, we know there are two positive real roots to this equation. From the graph, we know the other two roots are complex. The second positive real root is extraneous, corresponding to the negative branch of the square root function.