Question 28

The product of the real solutions $$x$$ of the equation
$$x^2 + 4 \mid x \mid - 4 = 0$$ is

Solution

$$x^2 + 4 \mid x \mid - 4 = 0$$

Case 1:

x > 0

$$x^2 + 4 x - 4 = 0$$

x = -2+2$$\sqrt{2}$$, -2-2$$\sqrt{2}$$

Since x > 0, x = -2+2$$\sqrt{2}$$

Case 2:

x < 0

$$x^2 - 4 x - 4 = 0$$

x = 2+2$$\sqrt{2}$$, 2-2$$\sqrt{2}$$

Since x < 0, x = 2-2$$\sqrt{2}$$

Product of values of x

=-2+2$$\sqrt{2}$$ * 2-2$$\sqrt{2}$$

= -12+8$$\sqrt{2}$$

= $$-4(\sqrt2 - 1)^2$$

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