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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