Hidden Quadratic Equations (C)

Sometimes quadratic equations come in disguise like

$$u^4 - 3u^2 + 4 = 0\;\;{\rm or}\;\;\frac{u + 2}{u^2 + 1} = 4$$

or even  

$$\frac{e^x - e^{-x}}{2} = 4$$ (2)

which can be seen to be a quadratic equation in e^x by multiplying through by 2e^x.

Hidden quadratic equations such as (2) are most easily solved by substitution. For example, in (2) x only appears as e^x (e^-x = 1/e^x), we let w = e^x to get  

$$\frac{w - (1/w)}{2} = 4.$$ (3)

Multiplying (3) through by 2w yields w² - 1 = 8w or w²-8w - 1 = 0, with the side condition that w > 0. Ignoring the side condition for the moment and applying the quadratic formula gives

$$w = \frac{8 \pm \sqrt{64 + 4}}{2} = \frac{8 \pm \sqrt{68}}{2} = 4 \pm \sqrt{17}.$$

Since w > 0 we have $e^x = w = 4 + \sqrt{17}$, so $x = \ln(4 + \sqrt{17})$.

Exercises In each case, solve for x. Remember, x could be a complex number.

1.

x⁴ + 2x² + 1 = 0;

2.

$${\frac{x^2 + 9x + 1}{x^2 + 5x + 3} = \frac{1}{3}}$$;

3.

$${\frac{x^2 + 9x + 1}{5x + 3} = 5x-3}$$;

4.

e^x + e^-x = 8.

5.

Find the length of the equal sides of an isosceles triangle if the area of the triangle is 5 and the remaining side has length 4.