The function is defined by , where , , and are constants, , and . The graph of in the -plane passes through the points and . Which choice is the value of in terms of ?

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Super-Hard SAT Math Question 182 | Free SAT Prep | Aniko

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Question 182·200 Super-Hard SAT Math Questions·Advanced Math

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The function $f$ is defined by $f(x)=ax^2+bx+c$ , where $a$ , $b$ , and $c$ are constants, $a>0$ , and $a\ne \frac12$ . The graph of $y=f(x)$ in the $xy$ -plane passes through the points $(1+\sqrt{2a},0)$ and $(1-\sqrt{2a},0)$ . Which choice is the value of $\frac{b^2}{c}$ in terms of $a$ ? © ani⁠kο.aі

When you’re given two points where $y=0$ on a quadratic, treat them as roots and write the function as $a(x-r_1)(x-r_2)$ . If the roots are conjugates (involving $\pm\sqrt{\,}$ ), multiply using $(u-v)(u+v)=u^2-v^2$ to avoid messy expansion. Then read $b$ and $c$ from the expanded form and simplify the target expression by canceling common factors. Рr⁠οpеrt‍y‍ o‌f Anіko.аі

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Start from the zeros

Because the points have $y=0$ , you can write $f(x)=a(x-r_1)(x-r_2)$ using their $x$ -coordinates as roots.

Multiply efficiently

Treat the factors as conjugates with $u=x-1$ and $v=\sqrt{2a}$ so you can use $(u-v)(u+v)=u^2-v^2$ . Writte⁠n bу Anikо

Simplify $\frac{b^2}{c}$ carefully

Once you identify $b$ and $c$ , factor out common powers of $a$ in $\frac{b^2}{c}$ to reduce the expression.

Desmos Guide

Define the function from its roots

Create a slider $a$ (and keep in mind $a>0$ ). Then enter

f(x)=a(x-(1+\sqrt{2a}))(x-(1-\sqrt{2a})).

Compute $c$ as $f(0)$

Enter

c=f(0).

Desmos will express $c$ in terms of $a$ .

Compute $b$ using $f(1)=a+b+c$

Enter ЅАT⁠ prе‌р bу Аnіko.⁠aі

b=f(1)-a-c.

This isolates $b$ because $f(1)=a+b+c$ for $f(x)=ax^2+bx+c$ .

Form $E=b^2/c$ and match a choice

Enter

E=b^2/c.

Compare Desmos’s simplified expression for $E$ to the answer choices and select the match.

Step-by-step Explanation

Write $f(x)$ from its zeros

Since $(1+\sqrt{2a},0)$ and $(1-\sqrt{2a},0)$ have $y=0$ , they are zeros of $f$ .

So,

f(x)=a\bigl(x-(1+\sqrt{2a})\bigr)\bigl(x-(1-\sqrt{2a})\bigr).

Use the conjugate pattern to expand

Rewrite the factors as conjugates:

\bigl(x-(1+\sqrt{2a})\bigr)\bigl(x-(1-\sqrt{2a})\bigr)=\bigl((x-1)-\sqrt{2a}\bigr)\bigl((x-1)+\sqrt{2a}\bigr).

Then apply $(u-v)(u+v)=u^2-v^2$ :

\begin{aligned} &\bigl((x-1)-\sqrt{2a}\bigr)\bigl((x-1)+\sqrt{2a}\bigr) =(x-1)^2-(\sqrt{2a})^2 \\ &=(x-1)^2-2a =x^2-2x+1-2a. \end{aligned}

f(x)=a\bigl(x^2-2x+1-2a\bigr)=ax^2+(-2a)x+a(1-2a).

Read off $b$ and $c$

Matching $f(x)=ax^2+bx+c$ gives

b=-2a\quad\text{and}\quad c=a(1-2a).

Compute $\frac{b^2}{c}$

\frac{b^2}{c}=\frac{(-2a)^2}{a(1-2a)}=\frac{4a^2}{a(1-2a)}=\frac{4a}{1-2a}.

Therefore, the correct choice is $\frac{4a}{1-2a}$ . ЅАT prep ‍bу Aniкo.⁠aі

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation