Super-Hard SAT Advanced Math Question 12 | Free SAT Prep

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Question 12·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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Let $a$ , $b$ , $c$ , and $d$ be positive integers. The equation below is true for every value of $x$ .

\begin{aligned} \left(x^2+ax+b\right)^2-\left(cx+d\right)^2 &=x^4+6x^3+3x^2 \\ &\quad -18x-11 \end{aligned}

Which choice is equal to $ad+bc$ ?

For polynomial identities, avoid fully expanding everything at once. Match coefficients beginning with the highest powers, because those often determine a variable immediately. Then use any given restrictions, such as positive integers, to rewrite the remaining variables efficiently. Finish by using the constant term, which often supplies the final condition needed to identify the values.

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Match coefficients

Since the equation is true for every $x$ , the coefficients of the same powers of $x$ must match on both sides.

Start with the highest useful power

Find $a$ by comparing the coefficients of $x^3$ . Then compare the coefficients of $x^2$ .

Use the integer condition

The $x^2$ coefficient equation shows that $c^2$ is even. Write $c=2k$ for a positive integer $k$ before using the $x$ and constant coefficients.

Desmos Guide

Use algebra first

Coefficient comparison is the faster main method. After writing $c=2k$ , the constant-term condition simplifies to $4k^4-21k^2+20=0$ .

Graph the equation in $k$

Enter $y=4x^4-21x^2+20$ . Identify the positive integer $x$ -intercept; it gives the value of $k$ .

Verify the constants

Use the calculator to enter $a=3$ , $b=2k^2-3$ , $c=2k$ , and $d=3k$ . Then evaluate $ad+bc$ to verify the result.

Step-by-step Explanation

Compare the leading coefficients

Expand only the parts needed for each coefficient. The coefficient of $x^3$ on the left is $2a$ , so

2a=6

Thus, $a=3$ .

Use the $x^2$ and $x$ coefficients

The coefficient of $x^2$ gives

9+2b-c^2=3

so $c^2-2b=6$ . Therefore, $c$ is even. Let $c=2k$ , where $k$ is a positive integer. Then

b=2k^2-3

The coefficient of $x$ gives

6b-2cd=-18

Substituting $b=2k^2-3$ and $c=2k$ shows that $d=3k$ .

Use the constant term

The constant term on the left is $b^2-d^2$ , so

\begin{aligned} (2k^2-3)^2-(3k)^2&=-11 \\ 4k^4-21k^2+20&=0 \\ (4k^2-5)(k^2-4)&=0 \end{aligned}

Since $k$ is a positive integer, $k^2=4$ , so $k=2$ .

Find the requested expression

With $k=2$ ,

a=3,\quad b=5,\quad c=4,\quad d=6

Therefore,

ad+bc=(3)(6)+(5)(4)=38

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