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 $p+w$ ?

A quadratic with exactly one real solution has a discriminant of zero. Use that condition to connect its coefficients, then find the single root and apply the requirement that it be an integer. Desmos can calculate the remaining possible sums. The trap is stopping at one real root: that root might be a fraction.

Hints

Hint 1
The discriminant is $b^2-4ac$ for $ax^2+bx+c=0$ . It must be $0$ for exactly one real solution. What equation does that give you for $(p-4)w$ ?
Hint 2
When the discriminant is $0$ , the repeated root is $-b/(2a)$ . Here $a=p-4$ and $b=-12$ . Which positive integer values of $p-4$ make that root an integer?
Hint 3
Let $a=p-4$ . The coefficient relation gives $w=36/a$ , while $p=a+4$ . Keep each factor pair together when you calculate $p+w$ .

Step-by-step

Use the discriminant and the integer root
Step 1
Turn one real solution into an equation
The discriminant of $ax^2+bx+c=0$ is $b^2-4ac$ . It equals $0$ when the quadratic has exactly one real solution. Here $a=p-4$ , $b=-12$ , and $c=w$ . Substitute those coefficients: $(-12)^2-4(p-4)w=0$ . Square $-12$ : $144-4(p-4)w=0$ .
Step 2
Isolate the product
Add $4(p-4)w$ to both sides: $144=4(p-4)w$ . This puts the two unknown factors together.
Step 3
Find the factor pairs to consider
Divide by $4$ : $36=(p-4)w$ . Since $p>4$ and $p,w$ are positive integers, both factors on the right are positive integers.
Step 4
Find the single root
With discriminant $0$ , the repeated root, or single solution, is $-b/(2a)$ . Substitute $a=p-4$ and $b=-12$ : $x=\frac{-(-12)}{2(p-4)}$ . Cancel the $2$ : $x=\frac{6}{p-4}$ . The integer condition now restricts $p-4$ .
Step 5
Restrict the possible values of p
Let $a=p-4$ . Because $p>4$ , $a$ is a positive integer. The single root $6/a$ must also be an integer, so $a$ must divide $6$ evenly. Its positive divisors are $a\in\{1,2,3,6\}$ .
Step 6
Calculate the possible sums
Since $p=a+4$ and $w=36/a$ , type $a=[1,2,3,6]$ and then $a+4+36/a$ in Desmos. It shows $[41,24,19,16]$ , in the same order as the values of $a$ . The sum $19$ comes from $a=3$ , giving $p=7$ , $w=12$ , and the integer root $x=2$ . So $p+w=19$ . Choice C.

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

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