Solving for x by completing the square in a problem where the solution doesn't seem to have symmetrical answers

Question

So I've been given this problem: $-14x^2 + 45x + 14 = 0$

And I've tried it a number of times but can't seem to solve it. The answer is supposed to be found by completing the square, and the solution is $(2+7x)(7-2x)$.

I end up with something along the lines of: $(x - \frac{45}{28} )^2 - \frac{1241}{784}$

Now we can take $a = (x - \frac{45}{28})$ and $b = \frac{1241}{784}$, so that we can use the formula $a^2 - b^2 = (a+b)(a-b)$, but I really don't see how that even approaches the given solution of $(2+7x)(7-2x)$.

Answer 1

One method of completing the square is to apply the formula $(a+b)(a-b)=a^2-b^2$ twice (once in each direction).

$$-14x^2+45x+14=0$$ $$x^2-\frac{45}{14}x-1=0$$ $$x\left(x-\frac{45}{14}\right)-1=0$$ $$\left(x-\frac{45}{28}+\frac{45}{28}\right)\left(x-\frac{45}{28}-\frac{45}{28}\right)-1=0$$ $$\left(x-\frac{45}{28}\right)^2-\left(\frac{45}{28}\right)^2-1=0\tag{1}$$ $$\left(x-\frac{45}{28}\right)^2-\frac{2025}{784}-1=0$$ $$\left(x-\frac{45}{28}\right)^2-\frac{2809}{784}=0$$ $$\left(x-\frac{45}{28}\right)^2-\left(\frac{53}{28}\right)^2=0$$ $$\left(x-\frac{45}{28}+\frac{53}{28}\right)\left(x-\frac{45}{28}-\frac{53}{28}\right)=0\tag{2}$$ $$\left(x+\frac27\right)\left(x-\frac72\right)=0$$

Do you see how the formula was applied to get $(1)$ and $(2)$? The answer still doesn't look like the one they gave us, though.... Ah! But remember that we divided everything by $-14$? So, now, we can multiply by $-14$ to get $$-14\left(x+\frac27\right)\left(x-\frac72\right)=0$$ $$7\cdot(-2)\cdot\left(x+\frac27\right)\left(x-\frac72\right)=0$$ $$7\cdot\left(x+\frac27\right)\cdot(-2)\cdot\left(x-\frac72\right)=0$$ $$(7x+2)(-2x+7)=0$$ $$(2+7x)(7-2x)=0$$

Answer 2

$-14x^2+45x+14=0$

$-14x^2+45x=-14$

$14x^2-45x=14$

$x^2-\frac{45}{14}x=\frac{14}{14}$

$x^2-\frac{45}{14}x=1$

$x^2-\frac{45}{14}x+(\frac{45}{28})^2=1+(\frac{45}{28})^2$

$(x-\frac{45}{28})^2=1+(\frac{45}{28})^2$

$x-\frac{45}{28}=+/-\sqrt{[1+(\frac{45}{28})^2]}$

$x=+/-\sqrt{[1+(\frac{45}{28})^2]}+\frac{45}{28}$

You could see http://www.purplemath.com/modules/sqrquad.htm It explains everything simply.

Answer 3

You are doing the calculations wrong.

$-14x^2+45x+14=0$

$x^2 - \frac{45}{14}x - 1 = 0$

$x^2 - \frac{45}{14}x + \frac{2025}{784} - 1 - \frac{2025}{784} = 0 $

Which is $\ \ (x - \frac{45}{28})^2 - \frac{2809}{784} = 0 \ $, and not what you got.

From there, you get $\ \ (x - \frac{45}{28})^2 - \frac{53^2}{28^2} = (x + \frac{8}{28})(x - \frac{98}{28}) \ \ $, which is the answer you were looking for.

Answer 4

Another way of completing the square is to take $ax^2+bx+c$ and multiply by $4a$ to obtain $$4a^2x^2+4abx+4ac=(2ax+b)^2+4ac-b^2$$

Here multiply by $-56$ to obtain $$56\cdot 14x^2-56\cdot 45x-56\cdot 14=(-28x+45)^2-2025-784=(28x-45)^2-2809$$

Then $2809=53^2$ so you get $28x-45=\pm 53$

Taking this approach saves the fractions. When you solve this you will find that you get the answers you are looking for.