Super-Hard SAT Math Question 84 | Free SAT Prep | Aniko

00:00

Question 84·200 Super-Hard SAT Math Questions·Advanced Math

0 / 200

\sqrt{a-x}=30-2x

In the given equation, $a$ is a constant. The equation has exactly one real solution. Which choice is the minimum possible value of $16a$ ?

When a parameter equation involves a square root equaling a linear expression, first enforce the nonnegativity condition on the linear side. Then use a substitution that makes the linear expression a single variable (like $u=30-2x$ ), square to get a quadratic, and use discriminant/sign reasoning to control the number of valid solutions (especially paying attention to restrictions like $u\ge 0$ ). Finally, compare the parameter values from the different “exactly one solution” cases to find the minimum requested expression.

Hints

Click me!

Use the fact that a square root is nonnegative

Since $\sqrt{a-x}\ge 0$ , the expression $30-2x$ must also be $\ge 0$ . Consider substituting a new variable for $30-2x$ .

Turn it into a quadratic

After substituting $u=30-2x$ , isolate the square root and square both sides to get a quadratic equation in $u$ .

Count solutions using discriminant and signs

To have exactly one real solution in $x$ , you need exactly one solution with $u\ge 0$ . Think about when a quadratic has (1) a double root or (2) one positive and one negative root.

Desmos Guide

Graph both sides with a slider

Enter $y=\sqrt{a-x}$ and $y=30-2x$ . Create a slider for $a$ .

Find when there is exactly one intersection

Move the slider $a$ downward from a large value. Watch the graphs: for some values there will be one intersection, then at a certain point it will switch to two intersections, and below that there will be none.

Identify the boundary case

Adjust $a$ to the boundary where the two-intersection case just turns into one intersection (the line and curve just touch). Click the intersection point to confirm there is only one.

Compute $16a$ from the slider value

Read the slider value of $a$ at that boundary and multiply by 16. Match that result to the answer choice.

Step-by-step Explanation

Substitute to control the square root condition

Because $\sqrt{a-x}\ge 0$ , the right side must satisfy $30-2x\ge 0$ .

Let $u=30-2x$ (so $u\ge 0$ and $x=\frac{30-u}{2}$ ). Then the equation becomes

\sqrt{a-\frac{30-u}{2}}=u.

Square and write a quadratic in $u$

Inside the square root,

a-\frac{30-u}{2}=a-15+\frac{u}{2}.

\sqrt{a-15+\frac{u}{2}}=u.

Squaring both sides (valid because both sides are nonnegative) gives

a-15+\frac{u}{2}=u^2.

Multiply by 2:

2a-30+u=2u^2

2u^2-u-(2a-30)=0.

Determine when there is exactly one valid (nonnegative) root

The discriminant is

D=(-1)^2-4(2)\bigl(-(2a-30)\bigr)=1+8(2a-30)=16a-239.

For at least one real solution, $D\ge 0$ , so $a\ge \frac{239}{16}$ .

Now consider the constant term of the quadratic in $u$ , which is $-(2a-30)$ :

If $a>15$ , then $2a-30>0$ , so $-(2a-30)<0$ . The product of the roots is negative, so one root is positive and the other is negative. Since we also require $u\ge 0$ , this produces exactly one valid $u$ .
If $a<15$ , then $2a-30<0$ , so $-(2a-30)>0$ . The product of the roots is positive and the sum of the roots is $\frac{1}{2}>0$ , so (when real) both roots are positive, giving two solutions. The only way to get exactly one solution in this case is if the roots merge into a double root, meaning $D=0$ .
If $a=15$ , then $2u^2-u=0$ gives $u=0$ and $u=\frac12$ , which are two solutions, so this case is not allowed.

Find the minimum possible value of $16a$

From the analysis, the equation has exactly one real solution when either:

$a>15$ (which would make $16a>240$ ), or
$D=0$ , which gives

16a-239=0 \quad\Rightarrow\quad 16a=239.

Since $239<240$ , the minimum possible value of $16a$ is 239.

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation