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.
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$\begingroup$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 $x=\frac{3}{4} + \frac{\sqrt{17}}{\sqrt{16}} = \frac{3}{4} + \frac{\sqrt{17}}{4} = \frac{3 + \sqrt{17}}{4}$
Another way of completing the square is to multiply rather than dividing. It goes like this:
$$ax^2+bx+c=0$$
Multiply by $4a$ to get $$4a^2x^2+4abx+4ac=(2ax+b)^2+4ac-b^2=0$$ Or $$(2ax+b)^2=b^2-4ac$$and the normal quadratic formula follows.
In your case the multiplier is $4a=4\times 2 =8$
So that $$16x^2-24x-8=(4x-3)^2-17=0$$
The marginal advantages are that it avoids using fractions before they are necessary, so the expressions come out less complicated, and that avoids errors. Also the first step can be done in any context in which multiplication makes sense (e.g. integral domains or rings of integers) - even if you can't take square roots or take multiplicative inverses.
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