Question 190·200 Super-Hard SAT Math Questions·Advanced Math

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For positive integers $a$ and $b$ with $b>2a$ , the equation

\sqrt{2x-b}=x-a

has exactly one real solution. If that solution is $x=6$ , which choice could be the value of $a+b$ ?

In a radical equation with unknown constants, a known solution lets you relate those constants. The square root’s nonnegative value limits one constant; then Desmos can compare the resulting expressions to enforce a strict inequality. A strict inequality excludes its boundary points. After finding the constants, check the number of solutions: squaring a radical equation can introduce candidates that fail the original.

Hints

Hint 1
A square root is never negative. Substitute the given solution $x=6$ into the equation. What limit does the right side then place on the positive integer $a$ ?
Hint 2
Squaring the equation at $x=6$ gives a relationship between $a$ and $b$ . Combine it with the strict inequality $b>2a$ . On a graph, look for where one expression is strictly above the other.
Hint 3
Finding constants that make $x=6$ work is not enough. Graph both sides of the resulting radical equation and check whether they meet anywhere else.

Step-by-step

Narrow the integers, then check the solution count
Step 1
Use the given solution
The problem says $x=6$ is a solution, meaning it makes both sides equal. Substitute it into the equation:
$\sqrt{12-b}=6-a.$
Step 2
Limit the value of $a$
A square root is never negative, so $6-a$ cannot be negative. Since $a$ is a positive integer,
$1\le a\le6.$
Step 3
Relate the two constants
Square both sides of the equation from the first step: $12-b=(6-a)^2$ . This keeps the relationship that the given solution must satisfy.
Step 4
Bring in the strict inequality
Because $b>2a$ , subtract each side from $12$ , reversing the inequality: $12-b<12-2a$ . Replace $12-b$ using the previous step: $(6-a)^2<12-2a$ . The less-than sign is strict; equality won’t work.
Step 5
Find the allowed integer on a graph
In Desmos, let the horizontal coordinate $x$ stand for $a$ . Type $y=(6-x)^2$ and $y=12-2x$ . Click their crossings: $(4,4)$ and $(6,0)$ . The first graph is below the second only for $4<x<6$ . The only integer there is $a=5$ ; the boundary points $4$ and $6$ fail the strict inequality.
Step 6
Find the matching value of $b$
Return to $\sqrt{12-b}=6-a$ with $a=5$ . Type $\sqrt{12-x_1}\sim6-5$ in Desmos, using $x_1$ for the unknown $b$ . The $\sim$ tells Desmos to find the value that makes the sides equal. Under PARAMETERS, it shows $x_1=11$ , so $b=11$ .
Step 7
Graph both sides of the original equation
With $a=5$ and $b=11$ , type $y=\sqrt{2x-11}$ and $y=x-5$ in Desmos. An intersection of these new graphs is a solution of the original equation. Click where the two new graphs touch, at $(6,1)$ . They have no other intersection, so $x=6$ is the only real solution of the original radical equation.
Step 8
Answer what the question asks
The constants are $a=5$ and $b=11$ , so their sum is $a+b=5+11=16$ . Choice B.

Question 190·200 Super-Hard SAT Math Questions·Advanced Math

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