COMC 2008 QB2

Problem Statement

(a) Determine all real numbers $x$ such that $(x + 3)(x − 6) = −14$

(b) Determine all real numbers $x$ such that $2^{2x} − 3(2^x) − 4 = 0$

(c) Determine all real numbers $x$ such that $(x^2 − 3x)^2 = 4 − 3(3x − x^2)$

Problem Link

Solution

(a) $$(x + 3)(x − 6) = −14$$ $$x^2 - 3x - 4 = 0$$ $$(x - 4)(x + 1) = 0$$ $$x - 4 = 0 \implies x = 4$$ $$x + 1 = 0 \implies x = -1$$ (b) $$2^{2x} − 3(2^x) − 4 = 0$$ $$(2^x)^2 - 3(2^x) - 4 = 0$$ $$(2^x - 4)(2^x + 1) = 0$$ $$2^x - 4 = 0 \implies x = 2$$ $$2^x + 1 = 0 \implies \text{no solution}$$ (c) $$(x^2 − 3x)^2 = 4 − 3(3x − x^2)$$ $$ \text{Sub } y = x^2 - 3x: y^2 - 3y - 4 = 0$$ $$(y - 4)(y + 1) = 0$$ $$y - 4 = 0 \implies x^2 - 3x - 4 = 0$$ $$(x - 4)(x + 1) = 0 \implies x \in \{-1, 4\}$$ $$y + 1 = 0 \implies x^2 - 3x + 1 = 0$$ $$x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$$ $$x = \dfrac{3 \pm \sqrt{5}}{2}$$ $$\therefore x \in \{-1, 4, \dfrac{3 + \sqrt{5}}{2}, \dfrac{3 - \sqrt{5}}{2} \}$$