## Precalculus: Mathematics for Calculus, 7th Edition

$Find$ $all$ $real$ $solutions$ $of$ $the$ $quadratic$ $equation:$ $w^2 = 3(w-1)$ Distribute the 3 to $(w-1)$ $w^2 = 3w-3$ Subtract 3w from both sides $w^2-3w = 3w-3-3w$ $w^2-3w = -3$ Add 3 to both sides $w^2-3w+3 = -3+3$ $w^2-3w+3=0$ Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$ $1w^2-3w+3$ $a = 1, b = -3, c = 3$ $x = \frac{-(-3) \pm \sqrt {(-3)^2- 4(1 \times 3)}}{2(1)}$ $x = \frac{3 \pm \sqrt {9- 12}}{2}$ $x= \frac{3 \pm \sqrt {-3}}{2}$ You can never square root a negative number, so there is no real solution possible No Real Solution