A quadratic function is defined by . The graph of passes through the points and . If the equation has exactly one real solution, which choice is the least possible value of ?

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Super-Hard SAT Math Question 128 | Free SAT Prep | Aniko

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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 $a$ ? Aniko.⁠aі - ЅА‍T P‌rеp

When a quadratic is constrained to have exactly one real solution, immediately think “discriminant equals 0.” If additional information gives values of the function at specific $x$ -coordinates, use those points to build a small linear system to express $b$ and $c$ in terms of $a$ (or another parameter). Then substitute into $b^2-4ac=0$ to get a single equation in one variable and solve, being careful to apply the final instruction (here, choose the least value). Аnik‌ο Que‌ѕtiоn‌ Ваn⁠k

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Turn the points into equations

Use $f(1)=4$ and $f(2)=7$ to write two equations involving $a$ , $b$ , and $c$ .

Eliminate a variable efficiently

Subtract the two equations to eliminate $c$ and solve for $b$ in terms of $a$ . аniko.ai SAТ Q‍uеs⁠tion Ва‍nк

Use the condition on the number of real solutions

“Exactly one real solution” for a quadratic means its discriminant satisfies $b^2-4ac=0$ .

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Create a slider for $a$

Enter a=1 and make it a slider.

Define $b$ and $c$ from the point conditions

Enter b=3-3a and c=1+2a.

Define the discriminant as an expression in $a$

Enter D=b^2-4*a*c.

Find where the discriminant is zero

Graph y=D and find the $x$ -intercepts of this graph (these are the $a$ -values that make $D=0$ ). Choose the smaller intercept as the least possible value.

Step-by-step Explanation

Use the points to relate $a$ , $b$ , and $c$

Since $f(1)=4$ and $f(2)=7$ :

\begin{aligned} a+b+c &= 4 \\ 4a+2b+c &= 7 \end{aligned}

Subtract the first equation from the second:

(4a+2b+c)-(a+b+c)=7-4 \Rightarrow 3a+b=3

So $b=3-3a$ .

Write $c$ in terms of $a$

Substitute $b=3-3a$ into $a+b+c=4$ :

\begin{aligned} a+(3-3a)+c &= 4 \\ -2a + c &= 1 \\ c &= 1+2a \end{aligned}

Use the “exactly one real solution” condition

The equation $f(x)=0$ has exactly one real solution when the discriminant is 0: Po‍wеrеd bу А‌nikо

b^2-4ac=0

Substitute $b=3-3a$ and $c=1+2a$ :

(3-3a)^2-4a(1+2a)=0

Simplify:

\begin{aligned} 9(1-a)^2-4a-8a^2 &= 0 \\ 9(1-2a+a^2)-4a-8a^2 &= 0 \\ 9-18a+9a^2-4a-8a^2 &= 0 \\ a^2-22a+9 &= 0 \end{aligned}

Solve for $a$ and choose the least value

Solve $a^2-22a+9=0$ :

\begin{aligned} a &= \frac{22\pm\sqrt{22^2-4\cdot 1\cdot 9}}{2} \\ &= \frac{22\pm\sqrt{484-36}}{2} \\ &= \frac{22\pm\sqrt{448}}{2} \\ &= \frac{22\pm 8\sqrt{7}}{2} \\ &= 11\pm 4\sqrt{7} \end{aligned}

The least possible value is $11-4\sqrt{7}$ .

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