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