## Being Receptive- A Quadratic Example

A few days ago, I presented some students with this problem from the 2003 AMC 10A (problem #5):

Let $d$ and $e$ denote the solutions of $2x^2 + 3x - 5 = 0$. What is the value of $(d-1)(e-1)$?

Almost as soon as the problem was stated, the students began working out the solution by applying the quadratic formula to solve for $d$ and $e$, and then evaluating the desired formula by substituting their found values for $d$ and $e$.

I picked this problem, however, because it invites many different approaches.

Before addressing the most direct way, I want to describe a way to solve this problem using Vieta’s formulas. If you have a quadratic equation of the form $ax^2 + bx + c = 0$ and $r_1$ and $r_2$ are the roots, Vieta’s formulas tell us that $r_1 + r_2 = -\frac{b}{a}$ and $r_1r_2 = \frac{c}{a}$. This is readily verified by comparing coefficients in the equality $ax^2 + bx + c = a(x-r_1)(x-r_2)$.

Applying Vieta’s formulas to the specific problem, we can read off from the coefficients that $d + e = -1.5$ and $de = -2.5$. We can then evaluate $(d-1)(e-1)$ without ever applying the quadratic formula: $(d-1)(e-1) = de - (d+e) + 1 = -2.5 + 1.5 + 1 = 0$.

But there’s more.

Vieta’s formulas tell us that we can read off the product of the roots straight from the coefficients of the quadratic: the product is the ratio of the constant term to the leading coefficient. In this problem, we’re asked to find $(d-1)(e-1)$, which isn’t the product of the roots of the given quadratic. However, it is the product of the roots of the quadratic whose roots are $d-1$ and $e-1$, and a quadratic whose roots are $d-1$ and $e-1$ can be obtained from the given quadratic via a horizontal shift:

$2(x+1)^2 + 3(x+1) - 5 = 2x^2 + 4x + 2 + 3x + 3 - 5 = 2x^2 + 7x$,

and this has constant term zero. Hence, the answer is zero.

Even more, finding the constant term of a quadratic (or any polynomial) is simply a matter of evaluating the polynomial at zero. So we didn’t actually have to carry out the polynomial multiplication in the previous paragraph. Instead, we can just substitute $x = 0$ in before simplifying to find the constant term: $2(1)^2 + 3(1) - 5 = 0$ (and, if necessary, divide by the leading coefficient which will remain equal to 2).

But what does it mean to substitute $x = 0$ into the quadratic obtained by shifting the given quadratic horizontally to the left by 1? It’s the same as substituting $1$ into the original quadratic!

And so we come to perhaps the most direct way of seeing the answer which you might have noticed all along: When you have a quadratic equation of the form $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$, it means that $ax^2 + bx + c = a(x-r_1)(x-r_2)$. So in the problem, since we are told that $d$ and $e$ are the roots, we know that $2x^2 + 3x - 5 = 2(x-d)(x-e)$. Now, look carefully at the desired expression in the problem: $(d-1)(e-1)$. It’s almost equal to $2(x-d)(x-e)$ when $x = 1$; we just have to divide by 2 to get the desired number. Thus, the answer can be obtained by substituting $x=1$ into the given quadratic and then dividing by the leading coefficient, that is, the answer is one-half of $2+3-5=0$, or zero.

After going through all this, one student exclaimed, “I still prefer using the quadratic formula!” Well, okay, for those of you who would still prefer to use the quadratic formula, here’s a problem:

Let $d$ and $e$ denote the solutions of $3x^2 + 34x - 13 = 0$. What is the value of $(d-1)(e-1)$?

You go ahead and use the quadratic formula to solve that. Meanwhile, I’ll use the more direct method: “To get the answer, I just substitute $x = 1$ into the given quadratic and divide by the leading coefficient. So the answer is 3 + 34 – 13 = 24 divided by 3, which is 8. Done.”

Wake me up when you’ve finished solving it using the quadratic formula.

OK, just kidding. I really have nothing against using the quadratic formula to solve these problems. It’s a perfectly valid method and you’ll gain more experience evaluating expressions with radicals by using it, and that can’t hurt. (Though it is a slower and more error-prone method and if you’re doing this during a math competition, the time you save could be valuable.)

But it’s good to be receptive to other methods and techniques. Some day, your favored method may literally be inapplicable and you may be forced to use an alternative. Well, there’s no better some day than today, so try this:

Let $d$, $e$, $f$, $g$, and $h$ denote the solutions of

$x^5 - 5x^4 + 7x^3 - 2x^2 + 3x - 5 = 0$.

What is the value of

$(d-1)(e-1)(f-1)(g-1)(h-1)$?

Here, you’re not going to be able to find the exact values of the roots to substitute into the desired expression. However, if you apply the ideas discussed in this blog post, then you’ll find that the answer is 1.

When you encounter new methods, don’t give them the cold shoulder. Get to know them, add them to your toolbox, and use them!