Quadratic formula - check my simplificaiton

Question

I am trying to solve this equation using the quadratic formula:

$$x^2 + 4x -1 = 0$$

I start by substituting the values into the quadratic formula:

$$x = {-(4) \pm \sqrt {(4)^2 - 4(1)(-1)} \over 2}$$

which becomes

$$x = {-4 \pm \sqrt{20} \over 2}$$

This is the answer the textbook that I am using gives but I would have thought I could have simplified this further to:

$$x = {-4 \pm \sqrt {(5)(2)(2)} \over 2}$$

which becomes

$$x = {-4 \pm 2 \sqrt 5 \over 2}$$

which becomes

$$x = -2 \pm \sqrt 5$$

Am I right and if so, why would the textbook not have simplified it further?

Answer 1

People often have trouble checking their answers if they aren't in exactly the same form.

By presenting the answer in the form

$$ x = {-4 \pm \sqrt{20} \over 2} $$

students who have at least applied the quadratic formula correctly will not have trouble checking the result. Also, seeing the result in this form might help someone who is having trouble recognize what to do.

Students who go on to simplify this are probably able to check their answer either way.

Of course, I'm speculating as to the actual motivation of the textbook authors. And I do not intend to agree or disagree with the premise that the answer should have been given as $x = -2 \pm \sqrt{5}$.

Answer 2

Note that

$$\frac{B+C}{A}=\frac{B}{A}+\frac{C}{A}$$

---

$$x=\frac{-4\pm 2\sqrt{5}}{2}=\frac{-4}{2}\pm\frac{2\sqrt{5}}{2}=-2\pm\sqrt{5}.$$

Answer 3

Note in this case using the quadratic formula is silly, as it is so easy to complete the square (where the quadatic formula comes from): $$ x^2+4x-1=(x+2)^2-4-1=0\iff (x+2)^2=5\iff x=-2\pm\sqrt 5.$$

Answer 4

The classical formula for the quadratic equation is

$$ax^2+bx+c=0\iff x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

But there are many circumstances where the quadratic coefficient is $1$ and the first degree coefficient has an explicit factor $2$, and it is worth to remember the simplified formula $$x^2+2bx+c=0\iff x=-b\pm\sqrt{b^2-c}.$$

---

In your case, $$x=-2\pm\sqrt{2^2-(-1)}.$$

Answer 5

Correct me if I'm wrong but I don't think you are wondering how the authors got from

$x = {-4 \pm \sqrt {(5)(2)(2)} \over 2}$

to

$x = -2 \pm \sqrt 5$

(if so, mathlove already answered that).

But rather (or at least I personally would wonder) why the authors start with $x = {-4 \pm \sqrt {(5)(2)(2)} \over 2}$ in the first place.

Obviously they want to reduce it. However what you did is a lot more straight forward. Perhaps if they did it this way it would be easier to understand:

$x = {-4 \pm \sqrt{20} \over 2}$

first distribute:

$x = {-4 \over 2} \pm {\sqrt{20} \over 2}$

$x = -2 \pm {\sqrt{20} \over 2}$

now factor what's under the square root, if possible, into one or more factors that are a complete square, so the square can be taken out. The only complete square among the factors of 20 is 4, so:

$x = -2 \pm {\sqrt{(4)(5)} \over 2}$

$x = -2 \pm {2 \sqrt{5} \over 2}$

$2 \over 2$ cancels, so ...

$x = -2 \pm \sqrt{5}$

hope that helps.