Question 23·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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For a positive constant $k$ , consider the equation $(x^2-6x)^2-(k+6)(x^2-6x)+6k=0$ . The difference between the greatest and least real solutions of the equation is $14$ . What is the value of $k$ ?

Your answer

A repeated expression in a fourth-degree equation is a cue for substitution: treat that expression as one variable and factor. Completing the square then shows that the solutions come in pairs around a shared center. Use the given gap to locate the outer pair, then solve for the constant in Desmos. Don’t mistake a value of the substitute for a value of $x$ .

Hints

Hint 1
The repeated expression $x^2-6x$ can be treated as one quantity. Let $u=x^2-6x$ and factor the resulting quadratic in $u$ before you try to find any $x$ -solutions.
Hint 2
Use completing the square on $x^2-6x=u$ . Adding $9$ to both sides gives a square centered at $x=3$ , so solutions occur in mirror pairs around $3$ .
Hint 3
The outer roots are half the given gap away from $3$ . Compare that distance with the roots from $u=6$ . If those roots cannot be outermost, which branch must supply an outer root?

Step-by-step

Factor, find the outer roots, then solve for k
Step 1
Factor the repeated expression
Use substitution to name the repeated expression: let $u=x^2-6x$ . The given equation becomes $u^2-(k+6)u+6k=0$ . Factor it:
$(u-6)(u-k)=0.$
Step 2
Keep both possible values of u
The zero-product rule says a product is zero when at least one factor is zero. So both branches matter: $u=6$ or $u=k$ . Put $x^2-6x$ back in for $u$ :
$x^2-6x=6\quad\text{or}\quad x^2-6x=k.$
Step 3
Find the center of the roots
Both branches have the form $x^2-6x=u$ . Completing the square makes their shared shape visible. Add $9$ to both sides: $x^2-6x+9=u+9$ . Rewrite the left side: $(x-3)^2=u+9$ . Every root has a mirror root the same distance from $3$ .
Step 4
Locate the greatest and least roots
By that symmetry, the outer roots are $3-d$ and $3+d$ for some distance $d$ . Their difference is $14$ : $2d=14$ . Divide by $2$ : $d=7$ . So the least root is $3-7=-4$ , and the greatest is $3+7=10$ .
Step 5
Identify which branch reaches an outer root
For the $u=6$ branch, $(x-3)^2=6+9=15$ . Its roots are $\sqrt{15}$ away from $3$ , less than $7$ because $15<7^2=49$ . So neither outer root comes from $u=6$ ; in particular, $x=10$ must come from $u=k$ .
Step 6
Solve for the constant
Put $x=10$ into the original equation. In Desmos, type $h(t)=t^2-6t$ to shorten its repeated part. Then type $(h(10))^2-(k_1+6)h(10)+6k_1\sim0$ , using $k_1$ for the unknown $k$ . The $\sim$ tells Desmos to fit that value; under PARAMETERS, it shows $k_1=40$ . The $u=k$ roots are $10$ and its mirror $-4$ , and the $u=6$ roots lie between them. Their outer gap is $14$ , so Grid in 40.

Question 23·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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