Question 14·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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A quadratic function $q$ is defined by

q(x)=ax^2+bx+c,

where $a$ is a positive integer. The function $q$ has two distinct integer zeros and satisfies the following conditions:

\begin{aligned} q(3)&>0 \\[1.4em] q(5)&<0 \\[1.4em] q(10)&>0 \\[1.4em] q(0)&=72 \\[1.4em] q(6)&=-12 \end{aligned}

What is the value of $b$ ?

Your answer

For an upward-opening quadratic, its zeros separate the positive outputs from the negative ones. Use the signs to locate the integer zeros, then write the function in factored form. An element-list regression in Desmos can use the two exact outputs to find the remaining numbers. The key trap is that a negative output is not a zero: a zero has output $0$ .

Hints

Hint 1
Because $a$ is positive, the quadratic opens upward: its outputs are negative between its two zeros and positive outside them. What does the change from $q(3)>0$ to $q(5)<0$ tell you about the first integer zero?
Hint 2
The second zero must lie after an input with a negative output and before one with a positive output. Use $q(6)=-12$ and $q(10)>0$ to narrow its location.
Hint 3
A zero at $4$ gives the factor $x-4$ . Call the other zero $s$ , so factored form uses the two zeros and the leading coefficient. How can the outputs at $0$ and $6$ determine both unknowns?

Step-by-step

Approach 1: Locate the zeros, then fit the function

Step 1
Find the first integer zero
Since $a$ is positive, $q$ opens upward and is negative between its two zeros. The output changes from positive at $3$ to negative at $5$ , so the first zero is strictly between those inputs. It must be $4$ because the zeros are integers.
Step 2
Bound the second zero
A negative output places its input between the zeros. Since $q(6)=-12$ is negative but $q(10)>0$ , the second zero $s$ satisfies $6<s<10$ . It could be $7$ , $8$ , or $9$ . In particular, $6$ is not a zero: its output is $-12$ , not $0$ .
Step 3
Write the function from its zeros
A zero makes the function equal $0$ , so a zero at $4$ gives a factor $x-4$ , and a zero at $s$ gives $x-s$ . Keep the given leading coefficient $a$ to write factored form: $q(x)=a(x-4)(x-s)$ .
Step 4
Use both exact outputs in Desmos
The given outputs mean $a(0-4)(0-s)=72$ and $a(6-4)(6-s)=-12$ . Type the two expressions as a list, with their outputs in the same order after $\sim$ . The $\sim$ tells Desmos to fit $a$ and $s$ ; under PARAMETERS it shows $a=2$ and $s=9$ . The second zero fits $6<s<10$ .
Step 5
Expose the middle term
To find $b$ , expose the term containing $x$ by multiplying the factors: $q(x)=a\bigl[x^2-(4+s)x+4s\bigr]$ . Multiply through by $a$ : $q(x)=ax^2-a(4+s)x+4as$ .
Step 6
Read the requested coefficient
In $ax^2+bx+c$ , $b$ is the number multiplying $x$ , so $b=-a(4+s)$ . Type $-a(4+s)$ below the regression. Desmos uses its fitted values and prints $-26$ . That's the middle coefficient the question asks for. Grid in -26.

Approach 2: Use integer factors instead

Step 1
Turn the output at zero into a product
From the zeros found above, $q(x)=a(x-4)(x-s)$ . Put in $0$ because the problem gives $q(0)=72$ : $72=a(0-4)(0-s)=4as$ . Divide by $4$ : $18=as$ . This connects the integer zero $s$ to the positive integer $a$ .
Step 2
Apply the integer restriction
The sign conditions left $s$ equal to $7$ , $8$ , or $9$ . Since $as=18$ and $a$ is an integer, $s$ must divide $18$ evenly. Only $9$ does, so $s=9$ and $a=2$ .
Step 3
Use the sum of the zeros
For zeros $r$ and $s$ , expanding $a(x-r)(x-s)$ makes the middle coefficient $b=-a(r+s)$ . Here the zeros are $4$ and $9$ , so $b=-2(4+9)=-26$ . The minus sign comes from the factors $x-4$ and $x-9$ . Grid in -26.

Question 14·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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