# Using $x = y - b/2$ to solve a quadratic equation

I've been reading a book called Mathematics for the Nonmathematician, and it presents a solution to quadratic equations of the form:

$x^2 + bx + c = 0$

which relies on coming up with a new formula, namely:

$y = x + b/2$

which when rearranged gives:

$x = y - b/2$

which is then substituted into the original formula, giving:

$(y - b/2)^2 + b (y - b/2) + c = 0$

I understand that using $b/2$ makes it easier to simplify the equation.

However, the author also explained that when $b/2$ is added to both of the roots of an equation, the sum of the resulting values is equal to $0$. I am failing to understand the significance/proof of that. I'm not quite sure how the author has come to the conclusion that this will always be the case, and I'm not quite sure whether/why it is important.

I'm a complete maths novice, but a university course I am about to start has a mathematics component, and so I would like to understand as much as I can in an intuitive way before it begins.

• Interesting in its parallel to the quadratic equation, $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ However, I'm certain these types of problems have just been examined so much that ways often look the same or end up at the same point no matter which direction you're coming from. One way or another, you're going to provide a solution to the given equation and there are only so many relationships for such a "simple" shape. – jm324354 Jan 19 '15 at 22:46
• The proof is simple: If you expand the quadratic term the equation becomes $y^2 = b^2/4 - c$ with the two solutions $y = \pm \sqrt{b^2/4 - c}$ which sums to zero. – Winther Jan 19 '15 at 22:48
• I end up with $y^2 = b^2/2 - c$: $y^2 - by/2 - by/2 + by -b^2/2 + c$, leaving $y^2 - b^2/2 + c$. where am I going wrong? – subjectification Jan 19 '15 at 22:57
• The mistake is in the term: $(y-b/2)^2 = y^2 -2(b/2)y + (b/2)^2$. – Winther Jan 19 '15 at 23:03
• Ah, thank you very much. – subjectification Jan 19 '15 at 23:10

## 1 Answer

The equation $x^2 + bx + c = 0$ is derived from a more general form $ax^2 + bx + c = 0$

The curve $Y = ax^2 + bx + c$ has two characteristics:-

(1) The minimum occurs when $x = \frac {-b}{2a}$ (by assuming a>0);

(2) The axis of symmetry is the vertical line $X = \frac {-b}{2a}$.

Performing the substitution $x’ = x + \frac {-b}{2a}$ is just doing a transformation (translation actually) to the curve by shifting it horizontally left $\frac {-b}{2a}$ units.

After the transformation, the axis of symmetry is then $X = 0$. Such act can make the subsequent works (like roots finding [because they are now symmetrically located about the Y-axis]) a bit easier.

In your case, the ‘a’ is obviously equal to 1.

• That was a very helpful explanation, thank you. – subjectification Jan 20 '15 at 1:27
• @subjectification U r welcome. Hope that helps. – Mick Jan 21 '15 at 4:48