I have the following problem to solve using the method of completing the square.
$$2x^2-3x-1 = 0$$
Here is where I've gotten to so far on this problem.
$$2x^2-3x = 1$$
$$x^2-\frac{3}{2}x = \frac{1}{2}$$
$$x^2-\frac{3}{2}x +\frac{9}{16} = \frac{1}{2}+\frac{9}{16}$$
$$x^2-\frac{3}{2}x +\frac{9}{16} = \frac{17}{16}$$
$$(x -\frac{3}{4})^2 = \frac{17}{16}$$
The book that I'm using provides the answer, but not how the answer was arrived at. I can't figure out how to get from the last step that I completed to here:
$$x=\frac{3±\sqrt{17}}{4}$$
Thanks in advance for any assistance that you can provide.
Best Answer
You made a mistake in the last line. It should read $(x-\frac{3}{4})^2=\frac{17}{16}$...
This gives $(x-\frac{3}{4})^2 - \frac{17}{16} = 0$. Using the third binomial formula, we obtain $(x-\frac{3}{4} + \sqrt \frac{17}{16})(x-\frac{3}{4} - \sqrt \frac{17}{16}) = 0$.
Remembering that a product is zero iff one of its coefficients is zero, you have the answer.
Edit:
Consider $x=\frac{3}{4} + \sqrt \frac{17}{16}$ which is a solution to your equation.
Remember that for any two numbers $a,b$ $\sqrt{a b}=\sqrt{a} \sqrt{b}$. The same is true for quotients (change $b$ to $\frac{1}{c}$).
Then we get $x=\frac{3}{4} + \frac{\sqrt{17}}{\sqrt{16}} = \frac{3}{4} + \frac{\sqrt{17}}{4} = \frac{3 + \sqrt{17}}{4}$