1. Let the length be $x$. Then the width is $x - 3$. The area is $x(x - 3) = 54$. Solving $x^2 - 3x - 54 = 0$ gives $(x - 9)(x + 6) = 0$, so $x = 9$ or $x = -6$. Since length must be positive, length = 9 units, width = 6 units.
2. Let the number be $x$. Then $x^2 + 2x = 48$. Rearranging: $x^2 + 2x - 48 = 0$. Factoring: $(x + 8)(x - 6) = 0$, so $x = -8$ or $x = 6$.
3. Let the number be $x$. Then $x^2 - 3x = 10$. Rearranging: $x^2 - 3x - 10 = 0$. Factoring: $(x - 5)(x + 2) = 0$, so $x = 5$ or $x = -2$. Since the number is positive, $x = 5$.
4. Vertex form: $y = a(x - 3)^2 + 2$. Plug in point (-1, 10): $10 = a(-1 - 3)^2 + 2 \Rightarrow 10 = 16a + 2 \Rightarrow 8 = 16a \Rightarrow a = \frac{1}{2}$. Expand: $y = \frac{1}{2}(x^2 - 6x + 9) + 2 = \frac{1}{2}x^2 - 3x + \frac{9}{2} + 2 = \frac{1}{2}x^2 - 3x + \frac{13}{2}$. So $a = \frac{1}{2}$, $b = -3$, $c = \frac{13}{2}$.
Parent Tip: Review the logic above to help your child master the concept of quadratic equation word problems worksheet with answers.
