Algebra - Equations and Inequations

Solve the equation: $\;$ $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) = 15$

Given equation: $\;$ $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) = 15$

i.e. $\;$ $\left(x^2 - 5x + 4\right) \left(x^2 - 5x + 6\right) = 15$ $\;\;\; \cdots \; (1)$

Let $\;$ $x^2 - 5x = p$ $\;\;\; \cdots \; (2)$

Then $(1)$ becomes

$\left(p +4\right) \left(p + 6\right) = 15$

i.e. $\;$ $p^2 + 10p + 24 = 15$

i.e. $\;$ $p^2 + 10p + 9 = 0$

i.e. $\;$ $\left(p + 9\right) \left(p + 1\right) = 0$

i.e. $\;$ $p = -9$ $\;$ or $\;$ $p = -1$

When $\;$ $p = -9$, $(2)$ becomes

$x^2 - 5x = -9$

i.e. $\;$ $x^2 - 5x + 9 = 0$ $\;\;\; \cdots \; (3)$

Discriminant of $(3)$ is

$\Delta = \left(-5\right)^2 - 4 \times 1 \times 9 = = 25 - 36 = - 9 < 0$

$\implies$ $(3)$ does not have any real roots.

When $\;$ $p = -1$, $\;$ $(2)$ becomes

$x^2 - 5x = -1$

i.e. $\;$ $x^2 - 5x + 1 = 0$ $\;\;\; \cdots \; (4)$

Solving $(4)$ for $x$ gives

$x = \dfrac{5 + \sqrt{21}}{2}$ $\;$ or $\;$ $x = \dfrac{5 - \sqrt{21}}{2}$

$\therefore \;$ Solution of the given equation is

$x = \left\{\dfrac{5 + \sqrt{21}}{2}, \; \dfrac{5 - \sqrt{21}}{2} \right\}$

Algebra - Equations and Inequations

Solve the equation: $\;$ $\left(x - \sqrt{3}\right)^4 - 5 \left(x - \sqrt{3}\right)^2 + 4 = 0$

Given equation: $\;$ $\left(x - \sqrt{3}\right)^4 - 5 \left(x - \sqrt{3}\right)^2 + 4 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\left(x - \sqrt{3}\right)^2 = p$ $\;\;\; \cdots \; (2)$

Then $(1)$ becomes

$p^2 - 5p + 4 = 0$

i.e. $\;$ $\left(p - 4\right) \left(p - 1\right) = 0$

i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = 1$

When $\;$ $p = 4$, $\;$ $(2)$ becomes

$\left(x - \sqrt{3}\right)^2 = 4$

i.e. $\;$ $x - \sqrt{3} = \pm 2$

i.e. $\;$ $x = \sqrt{3} + 2$ $\;$ or $\;$ $x = \sqrt{3} - 2$

When $\;$ $p = 1$, $\;$ $(2)$ becomes

$\left(x - \sqrt{3}\right)^2 = 1$

i.e. $\;$ $x - \sqrt{3} = \pm 1$

i.e. $\;$ $x = \sqrt{3} + 1$ $\;$ or $\;$ $x = \sqrt{3} - 1$

$\therefore \;$ The solutions of $(1)$ are

$x = \left\{\sqrt{3} + 2, \sqrt{3} - 2, \sqrt{3} + 1, \sqrt{3} - 1 \right\}$