ABCD-is-a-side-square-1-B-F-and-E-are-collinear-FDE-is-a-right-triangle-with-hypotenuse-1-and-the-DE-cathetus-is-worth-x-What-is-the-value-of-x-Solve-with-algebra-

Question Number 68601 by Maclaurin Stickker last updated on 14/Sep/19

A B C D is a side square 1 .

B, F and E are collinear .

F D E is a right triangle with hypotenuse 1

and the D E cathetus is worth x .

What is the value of x ?

(Solve with algebra)

Commented by Maclaurin Stickker last updated on 14/Sep/19

Commented by mr W last updated on 14/Sep/19

B E = \sqrt{{(1 + x)}^{2} + 1^{2}} = \sqrt{x^{2} + 2 x + 2}

\frac{1}{x} = \frac{\sqrt{x^{2} + 2 x + 2}}{1 + x}

\frac{1 + 2 x + x^{2}}{x^{2}} = x^{2} + 2 x + 2

x^{4} + 2 x^{3} + 2 x^{2} = 1 + 2 x + x^{2}

x^{4} + 2 x^{3} + x^{2} - 2 x - 1 = 0

\Rightarrow x \approx 0.8832

(e x a c t v a l u e p o s s i b l e, b u t c o m p l i c a t e d)

Commented by MJS last updated on 14/Sep/19

exact value, just for the fun of it

x^{4} + 2 x^{3} + x^{2} - 2 x - 1 = 0

(x^{2} + (1 - \sqrt{2}) x + 1 - \sqrt{2}) (x^{2} + (1 + \sqrt{2}) x + 1 + \sqrt{2}) = 0

x_{1} = - \frac{1}{2} + \frac{\sqrt{2}}{2} - \frac{\sqrt{- 1 + 2 \sqrt{2}}}{2} < 0 \Rightarrow not valid

x_{2} = - \frac{1}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{- 1 + 2 \sqrt{2}}}{2} > 0 \Rightarrow is the answer

x_{3, 4} = - \frac{1}{2} - \frac{\sqrt{2}}{2} \pm \frac{\sqrt{1 + 2 \sqrt{2}}}{2} i \notin R

Commented by Maclaurin Stickker last updated on 14/Sep/19

Ha, ha! Thank you, sir!