Let u = x² - 1. If u² - 7u + 10 = 0, how many distinct real values of x satisfy this equation?

Solve the quadratic in $u$ first: $u^2 - 7u + 10 = (u-2)(u-5) = 0$, so $u = 2$ or $u = 5$. Now substitute back $u = x^2 - 1$. If $u = 2$, then $x^2 = 3$, so $x = \pm\sqrt{3}$. If $u = 5$, then $x^2 = 6$, so $x = \pm\sqrt{6}$. These four values are all real and all different, so there are 4 distinct real solutions, choice A. Choice B (2) counts only one value of $u$ or only the positive roots. Choice C (1) and choice D (0) ignore that each positive value of $x^2$ gives two values of $x$, one positive and one negative.