1. **Problem Statement:** Find the discriminant of each quadratic equation and determine the number and type of solutions. 2. **Discriminant Formula:** The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by: $$\Delta = b^2 - 4ac$$ - If $\Delta > 0$, there are 2 distinct real solutions. - If $\Delta = 0$, there is 1 real solution (a repeated root). - If $\Delta < 0$, there are 0 real solutions (complex roots). 3. **Step 1: Calculate discriminants and solution types:** **a.** $10x^2 = 0$ means $a=10$, $b=0$, $c=0$. $$\Delta = 0^2 - 4 \times 10 \times 0 = 0$$ - One real solution (repeated root). **b.** $2x^2 - 9x + 8 = 0$, $a=2$, $b=-9$, $c=8$. $$\Delta = (-9)^2 - 4 \times 2 \times 8 = 81 - 64 = 17$$ - $\Delta > 0$ and not a perfect square, so 2 irrational solutions. **c.** $4x^2 - 21x - 18 = 0$, $a=4$, $b=-21$, $c=-18$. $$\Delta = (-21)^2 - 4 \times 4 \times (-18) = 441 + 288 = 729$$ - $\Delta > 0$ and $729 = 27^2$ is a perfect square, so 2 rational solutions. **d.** $x^2 + 18x + 97 = 0$, $a=1$, $b=18$, $c=97$. $$\Delta = 18^2 - 4 \times 1 \times 97 = 324 - 388 = -64$$ - $\Delta < 0$, so 0 real solutions. 4. **Step 2: Solve each equation using the specified methods:** **Factoring:** Solve $4x^2 - 21x - 16 = 0$. - Here $a=4$, $b=-21$, $c=-16$. - Find two numbers that multiply to $4 \times (-16) = -64$ and add to $-21$. - These are $-24$ and $3$. - Rewrite middle term: $$4x^2 - 24x + 3x - 16 = 0$$ - Group: $$ (4x^2 - 24x) + (3x - 16) = 0$$ - Factor each group: $$4x(x - 6) + 1(3x - 16) = 0$$ - Notice $3x - 16$ does not match $x - 6$, so try factoring differently. - Alternatively, use quadratic formula or check for error in problem statement. **Square Roots:** Solve $2x^2 - \frac{19}{2} = 0$. - Isolate $x^2$: $$2x^2 = \frac{19}{2}$$ $$x^2 = \frac{19}{4}$$ - Take square root: $$x = \pm \sqrt{\frac{19}{4}} = \pm \frac{\sqrt{19}}{2}$$ - Given $V=39$ and $x= \pm 7\sqrt{2}$, this seems unrelated; possibly a different problem. **Complete the Square:** Solve $2x^2 - 9x + 8 = 0$. - Divide entire equation by 2: $$x^2 - \frac{9}{2}x + 4 = 0$$ - Move constant: $$x^2 - \frac{9}{2}x = -4$$ - Take half of coefficient of $x$, square it: $$\left(\frac{-9/2}{2}\right)^2 = \left(-\frac{9}{4}\right)^2 = \frac{81}{16}$$ - Add to both sides: $$x^2 - \frac{9}{2}x + \frac{81}{16} = -4 + \frac{81}{16}$$ $$\left(x - \frac{9}{4}\right)^2 = \frac{-64}{16} + \frac{81}{16} = \frac{17}{16}$$ - Take square root: $$x - \frac{9}{4} = \pm \frac{\sqrt{17}}{4}$$ - Solve for $x$: $$x = \frac{9}{4} \pm \frac{\sqrt{17}}{4}$$ **Quadratic Formula:** Solve $x^2 + 16x + 97 = 0$. - $a=1$, $b=16$, $c=97$. - Formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ - Calculate discriminant: $$\Delta = 16^2 - 4 \times 1 \times 97 = 256 - 388 = -132$$ - Since $\Delta < 0$, solutions are complex: $$x = \frac{-16 \pm \sqrt{-132}}{2} = \frac{-16 \pm i\sqrt{132}}{2} = -8 \pm i\sqrt{33}$$ **Final answers:** - a) One real solution: $x=0$ - b) Two irrational solutions: $x = \frac{9 \pm \sqrt{17}}{4}$ - c) Two rational solutions: $x = \frac{21 \pm 27}{8}$ (from quadratic formula) - d) No real solutions, complex: $x = -8 \pm i\sqrt{33}$