Question 116·200 Super-Hard SAT Math Questions·Advanced Math

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\sqrt{a+x}+\sqrt{a-x}=10

In the given equation, $a$ is a constant. The equation has exactly one real solution. Which choice is the value of $4a$ ?

The terms $a+x$ and $a-x$ give this radical equation symmetry: replacing $x$ with $-x$ swaps the roots but leaves their sum unchanged. A nonzero solution would come with a partner, so exactly one solution forces $x=0$ . Substitute that value, find $a$ with a Desmos regression, and check that no other $x$ works. Making $x=0$ work alone doesn't prove uniqueness.

Hints

Hint 1
Changing $x$ to $-x$ swaps the two roots without changing their sum. If a nonzero value works, its negative works too. What must the solution be if there is exactly one?
Hint 2
Put that value of $x$ into the original equation. You'll get an equation with only $a$ unknown, which Desmos can solve using a regression.
Hint 3
Finding a value of $a$ that makes one $x$ work isn't enough. Square the whole sum of the roots and find its greatest possible value. Can the sum reach $10$ anywhere else?

Step-by-step

Use symmetry, then check uniqueness
Step 1
Locate the only possible solution
Replacing $x$ with $-x$ swaps the two square roots, so it leaves their sum unchanged. But the equation has exactly one real solution. Every nonzero solution would bring a second solution, its negative. So the only possible solution is $x=0$ .
Step 2
Use the solution to make an equation for $a$
Substitute the only possible $x$ , $0$ , into the given equation:
$\sqrt a+\sqrt a=10.$
Combine the equal roots:
$2\sqrt a=10.$
Step 3
Find the constant in Desmos
Type $2\sqrt{a_1}\sim10$ , using $a_1$ as Desmos's name for the unknown constant. The regression reports $a_1=25$ under PARAMETERS, so $a=25$ . You still need to check that this value gives exactly one solution.
Step 4
Square the entire sum
With $a=25$ , square the whole left side. The middle term is twice the product of the roots, so don't drop it:
$\left(\sqrt{25+x}+\sqrt{25-x}\right)^2=(25+x)+(25-x)+2\sqrt{(25+x)(25-x)}.$
Combine the terms and multiply inside the remaining root:
$\left(\sqrt{25+x}+\sqrt{25-x}\right)^2=50+2\sqrt{625-x^2}.$
Step 5
Show that the sum reaches 10 only once
Since $x^2\ge0$ , bound the remaining root for every $x$ where both roots are real:
$\sqrt{625-x^2}\le25.$
Apply that bound to the squared sum:
$50+2\sqrt{625-x^2}\le50+2(25)=(2\sqrt{25})^2.$
Both original roots are nonnegative, so their sum cannot exceed $10$ . Equality requires $x^2=0$ , or $x=0$ . That value works by the equation in step 2, so there is exactly one solution.
Step 6
Find the requested value
Keep the regression and type $4a_1$ on the next Desmos line. It displays $100$ , so the requested value is $4a=100$ . Choice B.

Question 116·200 Super-Hard SAT Math Questions·Advanced Math

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