Question 128·200 Super-Hard SAT Math Questions·Advanced Math

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A quadratic function is defined by $f(x)=ax^2+bx+c$ . The graph of $f$ passes through the points $(1,4)$ and $(2,7)$ . If the equation $f(x)=0$ has exactly one real solution, which choice is the least possible value of $a$ ?

Two points on a quadratic give two equations for its coefficients. The cue exactly one real solution tells you to set the discriminant, $b^2-4ac$ , equal to $0$ . First express $b$ and $c$ in terms of $a$ , then graph the resulting equation in Desmos. That equation can have two valid values of $a$ , so compare them before choosing the least.

Hints

Hint 1
A point $(u,v)$ on a function's graph means $f(u)=v$ : put the point's first coordinate into the function and get its second coordinate out. What equations do the two points give you?
Hint 2
For a quadratic to have exactly one real solution, its discriminant $b^2-4ac$ must be $0$ . First use the point equations to write $b$ and $c$ in terms of $a$ .
Hint 3
After you get an equation involving only $a$ , let Desmos's $x$ represent $a$ and graph it. Check both $x$ -intercepts: the question asks for the least possible coefficient, not just any coefficient that works.

Step-by-step

Approach 1: Use the discriminant and graph the possible values
Step 1
Turn the points into equations
A point $(u,v)$ on the graph means $f(u)=v$ . Substitute each point's input into the given function:
$a(1)^2+b(1)+c=4,\qquad a(2)^2+b(2)+c=7.$
Simplify each equation:
$a+b+c=4,\qquad 4a+2b+c=7.$
Step 2
Eliminate c to find b
Subtract the first equation from the second so $c$ cancels:
$(4a+2b+c)-(a+b+c)=7-4.$
Combine like terms:
$3a+b=3.$
Subtract $3a$ :
$b=3-3a.$
Step 3
Find c in terms of a
Replace $b$ in the first point equation with $3-3a$ :
$a+(3-3a)+c=4.$
Combine the $a$ terms:
$-2a+3+c=4.$
Move $-2a+3$ to the right:
$c=4+2a-3.$
Combine constants: $c=1+2a$ . Both missing coefficients now depend only on $a$ .
Step 4
Make the discriminant zero
Exactly one real solution means the discriminant $b^2-4ac$ is $0$ . The two solutions in the quadratic formula become the same when the square root of that expression is $0$ . Substitute $b=3-3a$ and $c=1+2a$ :
$(3-3a)^2-4a(1+2a)=0.$
Expand each product:
$9-18a+9a^2-4a-8a^2=0.$
Combine like terms:
$a^2-22a+9=0.$
Step 5
Graph the equation for a
Let Desmos's $x$ stand for the coefficient $a$ , not the input of $f$ . Type $y=x^2-22x+9$ and click both $x$ -intercepts, the points where $y=0$ . Desmos shows approximately $(0.417,0)$ and $(21.583,0)$ . So there are two possible values of $a$ , and the smaller is near $0.417$ .
Step 6
Match the smaller intercept
Type the four choices on separate Desmos lines. In order, Desmos prints about $0.417$ , $21.583$ , $11.417$ , and $-1.583$ . Only the first two match the intercepts, and $11-4\sqrt7$ matches the smaller one. So the least possible value of $a$ is $11-4\sqrt7$ . Choice A.

Approach 2: Find the exact values with the quadratic formula
Step 1
Solve the coefficient equation exactly
Instead of matching decimals, apply the quadratic formula to $a^2-22a+9=0$ . Here the coefficient of $a$ is $-22$ :
$a=\frac{-(-22)\pm\sqrt{(-22)^2-4(1)(9)}}{2(1)}.$
Calculate under the square root:
$a=\frac{22\pm\sqrt{448}}{2}.$
Simplify $\sqrt{448}=8\sqrt7$ :
$a=\frac{22\pm8\sqrt7}{2}.$
Divide both terms in the numerator by $2$ :
$a=11\pm4\sqrt7.$
Step 2
Take the lesser exact value
Since $4\sqrt7$ is positive, subtracting it gives the lesser of the two values: $11-4\sqrt7$ . That is the least possible coefficient $a$ . Choice A.

Question 128·200 Super-Hard SAT Math Questions·Advanced Math

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