Question 1000·Hard·Math

0 / 2000

81x^2 + (81q - p)x - pq = 0

In the equation above, $p$ and $q$ are positive constants. The product of the two solutions to the equation can be written as $-k pq$ , where $k$ is a positive constant. What is the value of $k$ ?

If a quadratic asks for the product of its solutions rather than the solutions themselves, read it from the coefficients. For $ax^2+bx+c=0$ , the product is $c/a$ ; then compare that expression with the form the question gives. Don’t take the square root of $a$ because the equation contains $x^2$ . Use the full coefficient in the denominator.

Hints

Hint 1
The leading coefficient multiplies $x^2$ , while the constant term has no $x$ . Identify those two parts of the equation. The middle term isn’t needed to find the product of the solutions.
Hint 2
The root-product rule says the solutions multiply to $c/a$ in $ax^2+bx+c=0$ . Keep the minus sign on the constant term, and divide by the full leading coefficient, not its square root.
Hint 3
Set the product you found equal to the form $-kpq$ . Since $p$ and $q$ are positive, their product isn’t zero, so you can divide both sides by $-pq$ to find $k$ .

Step-by-step

Use the root-product rule
Step 1
Identify the two coefficients that matter
No Desmos needed. The product comes directly from the coefficients. Read the equation as $ax^2+bx+c=0$ . Its leading coefficient, the number multiplying $x^2$ , is $a=81$ . Its constant term, the part with no $x$ , is $c=-pq$ . The middle term isn’t needed for the product.
Step 2
Find the product of the solutions
Call the solutions $r$ and $s$ . Factoring the quadratic as $a(x-r)(x-s)$ makes its constant term $ars$ , so the root-product rule is $rs=c/a$ . Substitute the coefficients: $rs=\frac{-pq}{81}$ . Divide the constant term by the full coefficient of $x^2$ .
Step 3
Match the requested form
The question writes that same product as $-kpq$ . Set the two expressions equal: $-kpq=\frac{-pq}{81}$ . Divide by $-pq$ , which isn’t zero because $p$ and $q$ are positive: $k=\frac{1}{81}$ . So the value of the positive constant $k$ is $\frac{1}{81}$ . Choice C.

Lessons that teach this

SAT Advanced AlgebraIntermediateCoreConnect quadratic roots and coefficients

Question 1000·Hard·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this