Super-Hard SAT Advanced Math Question 19 | Free SAT Prep

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Question 19·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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(p-4)x^2-12x+w=0

The constants $p$ and $w$ are positive integers, and $p>4$ . The quadratic equation has exactly one real solution, and that solution is an integer. Which choice could be the value of $p+w$ ?

For a parameter-based quadratic with exactly one real solution, set the discriminant equal to $0$ immediately. Then use any added condition, such as an integer root, to restrict the parameter values further. Introducing a simpler variable for a repeated expression, such as $d=p-4$ , can turn the discriminant equation into a manageable factor-pair problem.

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Focus on the discriminant

A quadratic has exactly one real solution when its discriminant is $0$ . Let $d=p-4$ to make the coefficient of $x^2$ easier to use.

Use the repeated root

After setting the discriminant equal to $0$ , find the repeated root using $\frac{-b}{2a}$ . The problem says this root must be an integer.

Connect the conditions

The discriminant condition makes $d$ and $w$ factors of $36$ , while the integer-root condition requires $d$ to divide $6$ .

Desmos Guide

Use algebra first

Algebra is faster because the discriminant and integer-root conditions identify all possible parameter pairs. Desmos is most useful for verifying a pair after it is found.

Verify a valid pair

Enter $y=(7-4)x^2-12x+12$ . The graph touches the $x$ -axis at exactly one point, showing that the equation has one real solution.

Check the solution value

Click the point where the graph touches the $x$ -axis. Its $x$ -coordinate is an integer, confirming that this parameter pair meets both conditions.

Step-by-step Explanation

Use the one-solution condition

Let $d=p-4$ . Since $p>4$ , $d$ is a positive integer. The equation becomes

dx^2-12x+w=0

Exactly one real solution means the discriminant equals $0$ :

(-12)^2-4(d)(w)=0

So $144-4dw=0$ , which simplifies to $dw=36$ .

Use the integer-solution condition

When the discriminant is $0$ , the repeated solution is

x=\frac{-(-12)}{2d}=\frac{6}{d}

For this solution to be an integer, the positive integer $d$ must be a divisor of $6$ .

List the possible sums

The possible values of $d$ are $1$ , $2$ , $3$ , and $6$ . Since $p=4+d$ and $w=36/d$ , the possible sums are:

$d$	$p$	$w$	$p+w$
$1$	$5$	$36$	$41$
$2$	$6$	$18$	$24$
$3$	$7$	$12$	$19$
$6$	$10$	$6$	$16$

Therefore, the value that could equal $p+w$ is $19$ .

Study Hints

Hints

Click me!

Focus on the discriminant

A quadratic has exactly one real solution when its discriminant is $0$ . Let $d=p-4$ to make the coefficient of $x^2$ easier to use.

Use the repeated root

After setting the discriminant equal to $0$ , find the repeated root using $\frac{-b}{2a}$ . The problem says this root must be an integer.

Connect the conditions

The discriminant condition makes $d$ and $w$ factors of $36$ , while the integer-root condition requires $d$ to divide $6$ .

Desmos Guide

Use algebra first

Algebra is faster because the discriminant and integer-root conditions identify all possible parameter pairs. Desmos is most useful for verifying a pair after it is found.

Verify a valid pair

Enter $y=(7-4)x^2-12x+12$ . The graph touches the $x$ -axis at exactly one point, showing that the equation has one real solution.

Check the solution value

Click the point where the graph touches the $x$ -axis. Its $x$ -coordinate is an integer, confirming that this parameter pair meets both conditions.

Step-by-step Explanation

Use the one-solution condition

Let $d=p-4$ . Since $p>4$ , $d$ is a positive integer. The equation becomes

dx^2-12x+w=0

Exactly one real solution means the discriminant equals $0$ :

(-12)^2-4(d)(w)=0

So $144-4dw=0$ , which simplifies to $dw=36$ .

Use the integer-solution condition

When the discriminant is $0$ , the repeated solution is

x=\frac{-(-12)}{2d}=\frac{6}{d}

For this solution to be an integer, the positive integer $d$ must be a divisor of $6$ .

List the possible sums

The possible values of $d$ are $1$ , $2$ , $3$ , and $6$ . Since $p=4+d$ and $w=36/d$ , the possible sums are:

$d$	$p$	$w$	$p+w$
$1$	$5$	$36$	$41$
$2$	$6$	$18$	$24$
$3$	$7$	$12$	$19$
$6$	$10$	$6$	$16$

Therefore, the value that could equal $p+w$ is $19$ .

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation