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 $a$ ?

Points on the $x$ -axis give a quadratic’s roots: inputs that make its output $0$ . Use the root sum to find $b$ and the root product to find $c$ , then form the requested ratio. For roots written as $m+t$ and $m-t$ , the sum cancels $t$ and the product is a difference of squares. Remember that the root product is $c/a$ , not $c$ .

Hints

Hint 1
A point with second coordinate $0$ has an output of $0$ , so its first coordinate is a root. How can you use the two roots to find the middle coefficient?
Hint 2
The sum of the roots of $ax^2+bx+c$ is $-b/a$ . Add the two given first coordinates: the opposite radical terms cancel.
Hint 3
The product of the roots is $c/a$ . Multiply the first coordinates using $(1+t)(1-t)=1-t^2$ , then remember to multiply by $a$ to get $c$ .

Step-by-step

Use the sum and product of the roots
Step 1
Turn the points into roots
No Desmos needed. The root sum and product give exact expressions in $a$ ; choosing a number for $a$ in Desmos would not prove a formula for every $a$ . Both points have output $0$ , so their first coordinates are roots, inputs that make $f(x)=0$ :
$f(1+\sqrt{2a})=0,\qquad f(1-\sqrt{2a})=0.$
Step 2
Use the root sum
For $ax^2+bx+c$ , the roots add to $-b/a$ . Apply that to the two first coordinates:
$-\frac{b}{a}=(1+\sqrt{2a})+(1-\sqrt{2a}).$
Cancel the opposite radical terms:
$-\frac{b}{a}=2.$
Step 3
Find the middle coefficient
Multiply both sides by $-a$ to isolate the middle coefficient:
$b=-2a.$
Step 4
Use the root product
For $ax^2+bx+c$ , the roots multiply to $c/a$ . Apply that to the two first coordinates:
$\frac{c}{a}=(1+\sqrt{2a})(1-\sqrt{2a}).$
Use the difference of squares pattern, $(1+t)(1-t)=1-t^2$ :
$\frac{c}{a}=1-(\sqrt{2a})^2.$
Square the radical:
$\frac{c}{a}=1-2a.$
Step 5
Find the constant term
The root product is $c/a$ , not $c$ . Multiply both sides by $a$ to get the constant term:
$c=a(1-2a).$
Step 6
Form the requested ratio
Since $a>0$ and $a\ne\frac12$ , neither $a$ nor $1-2a$ is $0$ , so division by $c$ is allowed. Substitute $b$ and $c$ :
$\frac{b^2}{c}=\frac{(-2a)^2}{a(1-2a)}.$
Square the numerator:
$\frac{b^2}{c}=\frac{4a^2}{a(1-2a)}.$
Cancel one factor of $a$ :
$\frac{b^2}{c}=\frac{4a}{1-2a}.$
That is the requested value of $\frac{b^2}{c}$ . Choice D.

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

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