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Question 96·Hard·Nonlinear Functions

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The function $f$ is defined by

f(x)=(x-2)(x+1)(x-8).

If $f(t^2-7t)=0$ , where $t$ is a real number, what is the sum of all possible values of $t$ ?

Your answer

When a function is given in factored form and you see something like $f(\text{expression})=0$ , first identify the roots of the function itself, then set the inside expression equal to each root. This turns one complicated equation into several simpler ones. If the result is a set of quadratics and the question asks for the sum of all solutions, use the sum-of-roots shortcut ( $-b$ for $t^2+bt+c=0$ ) instead of solving each quadratic individually; finally add those sums together, checking you accounted for every factor/root.

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Think about what makes a product zero

If $(x-2)(x+1)(x-8)=0$ , what must be true about $x$ ? Apply that same idea when the input is $t^2-7t$ instead of $x$ .

Use a substitution to simplify

Try letting $y=t^2-7t$ . Then your equation becomes $f(y)=0$ . What values of $y$ make $f(y)=0$ ?

Turn each value of the substitution back into an equation in $t$

Once you know the possible values of $y$ , set $t^2-7t$ equal to each of them. You will get three quadratic equations in $t$ .

Use a shortcut for the sum of roots

You need the sum of all possible $t$ values, not the individual values. For a quadratic $t^2+bt+c=0$ , how can you get the sum of the roots directly from $b$ without solving?

Desmos Guide

Graph the substituted expression as a function of t

In Desmos, type y = x^2 - 7x. Here, x is playing the role of $t$ , and this graph represents $y = t^2 - 7t$ .

Graph the values that make f(x) zero

On the same graph, add the three horizontal lines: y = 2, y = -1, and y = 8. These correspond to the values that $t^2-7t$ can take when $f(t^2-7t)=0$ .

Find all intersection points

Click on each intersection point between the parabola y = x^2 - 7x and the three lines. Record the $x$ -coordinates of all intersection points; these are the possible values of $t$ .

Add the t-values

In Desmos, you can create a list of those $x$ -values, for example A = {a1, a2, a3, a4, a5, a6} using your recorded values, and then type sum(A) to see their total. That total is the sum of all possible values of $t$ .

Step-by-step Explanation

Rewrite the equation using a substitution

We are given

f(x)=(x-2)(x+1)(x-8)

and the condition $f(t^2-7t)=0$ .

Let $y=t^2-7t$ . Then the condition becomes $f(y)=0$ .

So we need the values of $y$ that make

(y-2)(y+1)(y-8)=0.

Find the possible values of the inside expression

For a product to be zero, at least one factor must be zero. So we set each factor equal to zero:

$y-2=0$ gives $y=2$
$y+1=0$ gives $y=-1$
$y-8=0$ gives $y=8$

Remember $y=t^2-7t$ , so these give three equations:

$t^2-7t=2$
$t^2-7t=-1$
$t^2-7t=8$ .

Write the three quadratic equations in standard form

Rewrite each equation as a quadratic equal to zero:

$t^2-7t-2=0$
$t^2-7t+1=0$
$t^2-7t-8=0$

Each of these equations has two real solutions for $t$ , so in total there are 6 possible values of $t$ .

Use sum-of-roots to avoid solving each quadratic

For a quadratic equation $t^2+bt+c=0$ , the sum of its roots is $-b$ .

Each of our equations has the form $t^2-7t+(\text{constant})=0$ , so in every case $b=-7$ , and the sum of the two roots is

-(-7)=7.

So:

The two solutions of $t^2-7t-2=0$ add to 7.
The two solutions of $t^2-7t+1=0$ add to 7.
The two solutions of $t^2-7t-8=0$ add to 7.

The sum of all six possible values of $t$ is

7+7+7=21.

So the sum of all possible values of $t$ is $\boxed{21}$ .

Question 96·Hard·Nonlinear Functions

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Step-by-step Explanation

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