Completing the square Quadratics

Question

Solve this quadratic equation by completing the square:

$2x^2+x-4=0$

Can I have the method aswell please.

Answer 1

$$\begin{align}2x^2+x-4&=2\left(x^2+\frac 12x\right)-4\\&=2\left(x^2+2\times \frac 14x+\left(\frac{1}{4}\right)^2-\left(\frac{1}{4}\right)^2\right)-4\\&=2\left(\left(x+\frac 14\right)^2-\frac{1}{16}\right)-4\\&=2\left(x+\frac 14\right)^2-\frac{2}{16}-4\\&=2\left(x+\frac 14\right)^2-\frac{33}{8}.\end{align}$$

Hence, $$\begin{align}2x^2+x-4=0&\iff 2\left(x+\frac 14\right)^2-\frac{33}{8}=0\\&\iff 2\left(x+\frac 14\right)^2=\frac{33}{8}\\&\iff \left(x+\frac 14\right)^2=\frac{33}{16}\\&\iff x+\frac 14=\pm\sqrt{\frac{33}{16}}\\&\iff x=-\frac 14\pm\frac{\sqrt{33}}{4}.\end{align}$$

Answer 2

$2x^2+x-4=0\iff x^2+\frac{1}{2}x=2\iff x^2+\frac{1}{2}x+(\frac{1/2}{2})^{2}=2+(\frac{1/2}{2})^{2}\iff (x+\frac{1}{4})^2=\frac{33}{16}$

$\iff x+\frac{1}{4}=\pm\frac{\sqrt{33}}{4}\iff x=-\frac{1}{4}\pm\frac{\sqrt{33}}{4}$.

Answer 3

First start by factoring out the $2$ $$\begin{array}{lll} 2x^2+x-4&=&0\\ 2(x^2+\frac{1}{2}x-2)&=&0\\ \end{array}$$ Now the problem is simplified to one of

Factor $x^2+\frac{1}{2}x-2$

The problem statement is illustrated graphically in figure 1. Notice that the purple blob (the constant term) has a negative area.

Figure 1

First, we split the middle term in half, leaving us with two $\frac{1}{4}x$ middle terms (see figure 2).

Figure 2

Next, we rearrange the pieces to almost form a square, as in figure 3.

Figure 3

Now here is where we complete the square. We need to add a small $\frac{1}{4}$ by $\frac{1}{4}$ square to the "almost" square to complete it. The completed square is shown in figure 4. Note, however that we need to add a negative square of area $-\frac{1}{16}$ so that we to total area unchanged.

Figure 4

Finally we merge the small square (with the area $-\frac{1}{16}$) with the blob (of area $-2$). We then take this new blob and reshape it into a square of negative area ($-2-\frac{1}{16}=-\frac{33}{16}$). This situation is shown in figure 5.

Figure 5

So now we have reduced the problem to a simple difference of squares, which we know how to factor. $$\begin{array}{lll} x^2+\frac{1}{2}x-2 &=& \bigg(x+\frac{1}{4}\bigg)^2 - \bigg(\frac{\sqrt{33}}{4}\bigg)^2\\ &=& \bigg((x+\frac{1}{4})+\frac{\sqrt{33}}{4}\bigg)\bigg((x+\frac{1}{4})-\frac{\sqrt{33}}{4}\bigg)\\ &=& \bigg(x-\frac{-1-\sqrt{33}}{4}\bigg)\bigg(x-\frac{-1+\sqrt{33}}{4}\bigg)\\ \end{array}$$ Applying this result to the original problem statement we have $$\begin{array}{lll} 2x^2+x-4&=&0\\ 2(x^2+\frac{1}{2}x-2)&=&0\\ 2\bigg(x-\frac{-1-\sqrt{33}}{4}\bigg)\bigg(x-\frac{-1+\sqrt{33}}{4}\bigg)&=& 0\\ \end{array}$$ Which solving for $x$ we have $$x\in\bigg\{\frac{-1-\sqrt{33}}{4}, \frac{-1+\sqrt{33}}{4}\bigg\}$$ For an alternative graphical explaination of a similar problem, see https://math.stackexchange.com/a/1004362/140156.

Answer 4

When you square a linear binomial $(x+a)^2$, you get the quadratic trinomial $x^2+2ax+a^2$.

Assume you want to match that to a general quadratic trinomial

$$Ax^2+Bx+C.$$

You first multiply by $A$ to yield $$Ax^2+2Aax+Aa^2.$$

Then you equate $B$ and $2Aa$, which gives

$$a=\frac B{2A}$$and $$A(x+a)^2=Ax^2+Bx+\frac{B^2}{4A}.$$

And you finish by correcting the independent term, so that

$$Ax^2+Bx+C=A\left(x+\frac B{2A}\right)^2+C-\frac{B^2}{4A}.$$

Under this form, it is possible to solve the quadratic equation, $$A\left(x+\frac B{2A}\right)^2+C-\frac{B^2}{4A}=0$$ using

$$\left(x+\frac B{2A}\right)^2=-\frac{C-\dfrac{B^2}{4A}}A.$$

You can fill the rest.