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Solve the following equations:
$\displaystyle{3x^2 - 4x - 2 = 0}$
$\displaystyle{z^2 + 8z - 3 = 0}$
$\displaystyle{2x^2 + 5x - 1 = 0}$
$\displaystyle{5x^2 = 13 - 2x}$
$\displaystyle{x(5x-2)=4}$
$\displaystyle{2x(x+2) = 3}$
$\displaystyle{x = \frac{2 \pm \sqrt{10}}{3}}$
$\displaystyle{z = -4 \pm \sqrt{19}}$
$\displaystyle{\frac{-5 \pm \sqrt{33}}{4}}$
$\displaystyle{\frac{-1 \pm \sqrt{66}}{5}}$
$\displaystyle{x = \frac{1 \pm \sqrt{21}}{5}}$
$\displaystyle{\frac{x = -2 \pm \sqrt{10}}{2}}$
Solve the following equations using all 4 methods (i.e., completing the square, Po-Shen Loh's method, the method of depressed terms, and the quadratic equation. If any solutions look different from one another, show they are actually the same solution.
$3x^2 + 6 = 10x$
$3t^2 + 8t + 3 = 0$
$5t^2 - 8t = 3$
$5m^2 + 3m = 2$
$x^2 - 6x + 3 = 0$
See Full Solutions (using the depressed terms method)
$\cfrac{5 \pm \sqrt{7}}{3}$
$\cfrac{-4 \pm \sqrt{7}}{3}$
$\cfrac{4 \pm \sqrt{31}}{5}$
$-1,\cfrac{2}{5}$
$3 \pm \sqrt{6}$
Solve the following equations.
$x - 3\sqrt{x} -4 = 0$
$x^{2/3} + x^{1/3} - 6 = 0$
$(2x-3)^2 - 5(2x-3) + 6 = 0$
$(2t^2 + t)^2 - 4(2t^2 + t) + 3 = 0$
$\displaystyle{x^4 - 9 = -2x^2}$
$\displaystyle{4 \left( \sqrt{\frac{x+1}{2}} + \frac{\sqrt{2x-4}}{4} \right) = 5\sqrt{2}}$
$\displaystyle{\frac{x}{x+2} + \frac{2}{x-3} = \frac{10}{x^2 - x - 6}}$
$2^x + 2^{-x} = 3$
$16$
$-27,8$
$\frac{5}{2},3$
$-\frac{3}{2},-1,\frac{1}{2},1$
$\displaystyle{x = \pm \sqrt{-1 + \sqrt{10^{\phantom{1}}}}}$
$\displaystyle{x = 3}$
no solutions
$\displaystyle{x = \log_2 \left(\cfrac{3 \pm \sqrt{5}}{2}\right)}$
For each function, use completing the square to find a formula for the function with only a single occurence of the variable. Then draw the graph of the function, indicating the coordinates of any interesting points (as always).
$f(x)=x^2+10x+20$
$g(x)=x^2-6x-5$
$h(x)=3x^2-7x-10$
$q(x)=-2x^2+4x-1$
$r(x)=x^2+x+1$
Click for full solutions to parts (a), (b), and (c)
The graphs for parts (d) and (e) can be found with Desmos.
Note, in part (d) the $x$-intercepts are $\left(\frac{2 \pm \sqrt{2}}{2},0\right)$, the $y$-intercept is at $((0,-1)$, and $(0,0)$ on the graph of $y=x^2$ moved under the transformations involved to $(1,1)$. This special point for quadratic functions is called the vertex of the graph.
In part (e), there are no $x$-intercepts, the $y$-intercept is at $(0,1)$, and the vertex is at $(-\frac{1}{2},\frac{3}{4})$.