Find constants in equivalent expressions

Some SAT questions say that two expressions are equal for all values of $x$ and hide one or two unknown constants inside them. Finding those constants by hand means expanding every product and matching the terms one power of $x$ at a time. It's easy to drop a sign along the way.

Here's a typical question.

Fast Desmos solution

For all values of $x$ means the two sides give the same number whatever $x$ is. So hand Desmos a whole list of $x$-values, and ask for the $p$ and $q$ that make the sides match at every one of them:

x_1=[1...10]

(2x_1+p)(x_1-3)-(x_1+q)(x_1+1)~x_1^2-10x_1-11

Desmos reports $p=2$ and $q=5$, so

By hand, you'd expand both products and match three pairs of coefficients. Here Desmos does all of that in two short lines.

The fit takes two lines.

1. Make a list of inputs. Type x_1=[1...10]. Square brackets make a list, and the three dots fill in every whole number from $1$ to $10$.
2. Fit the two sides. Type the left side with x_1 wherever the question has $x$, then the tilde, ~, where the equals sign would go, then the right side. The tilde tells Desmos to find the constants that make the two sides match.

Desmos then shows the constants it found under the regression line.

Why the list matters. An identity has to hold at every input, and each input gives Desmos one more equation the constants must satisfy. With ten inputs, the constants have to make all ten equations work, and that pins them down in almost every SAT question. You want more inputs than the highest power of $x$, and ten covers any SAT identity you're likely to meet.

Without the list, Desmos has nothing telling it that x_1 is a set of inputs. It treats x_1 as one more unknown, finds values that make the two sides match at a single point, and reports constants that don't work for every $x$.

As in the last lesson, check that no earlier line gives $p$ or $q$ a value. If one does, Desmos treats that letter as fixed and won't fit it.

A regression always reports numbers, even when no constants can make the two sides match. It gives the closest fit it can find. So before you answer, check that the fit is exact.

Type the left side minus the right side on a new line, still with x_1. Desmos shows the difference at every input. If every entry is $0$, the two sides really are equal for all ten inputs, and the constants are right. A tiny entry like $1.8\times10^{-15}$ counts as $0$ too: it's the calculator rounding in its last digits, not a real gap. For the opening question, the line

(2x_1+p)(x_1-3)-(x_1+q)(x_1+1)-(x_1^2-10x_1-11)

shows ten zeros.

Here's a case where the check catches a problem. In the equation $2(kx+3)=4x+7$, $k$ is a constant. Is there a value of $k$ that gives infinitely many solutions?

The fit 2(kx_1+3)~4x_1+7 reports $k\approx2.07$. That alone looks like an answer. But the check line shows entries like $-0.86$ and $0.43$, not zeros. No value of $k$ works, and you can see why: matching the $x$ terms needs $2k=4$, so $k=2$, but then the numbers without $x$ are $6$ and $7$, and they never match. Some SAT questions offer there is no such value as a choice for exactly this reason.

The check also settles decimals. A value like $2.9999999$ is Desmos rounding, so type $3$ in place of the letter in the check line. If it still shows zeros, $3$ is exact.

One limit: this "no such value" reading is safe when the constants are only multiplied by numbers, as $k$ is here. When a constant is squared or multiplied by another one, a nonzero check can also mean Desmos settled on the wrong fit, so match one coefficient by hand before you answer no such value.

When the identity has two fits

Squares hide a sign. Suppose $(ax+3)^2$ is equivalent to $16x^2+bx+9$ for all $x$, where $a$ and $b$ are constants and $b<0$. Expanding gives $a^2x^2+6ax+9$, so $a^2=16$, and $a$ can be $4$ or $-4$. Both make an identity: $a=4$ with $b=24$, and $a=-4$ with $b=-24$.

Desmos reports only one of them, and here it reports $a=4$ and $b=24$. That breaks the condition $b<0$. Add the condition to the end of the regression line in curly braces:

(ax_1+3)^2~16x_1^2+bx_1+9{b<0}

Now Desmos reports $a=-4$ and $b=-24$. So when a question adds a condition like $b<0$, $a>0$ or $p<q$, check the fit against it, and put it in braces if Desmos lands on the other fit.

When only a product is fixed

Sometimes two constants only ever appear multiplied together. Say $f(x)=ax+3$ and $g(x)=bx$, and $f(g(x))=12x+3$ for all $x$. Then $f(g(x))=a(bx)+3=abx+3$, so the identity says $ab=12$ and nothing more. The pairs $a=6,\ b=2$ and $a=12,\ b=1$ both work.

The fit a(bx_1)+3~12x_1+3 reports $a=12$ and $b=1$, one pair out of many. If the question asks for $ab$, type a*b and you get $12$, which is right for every pair. Don't answer with $a$ or $b$ alone. If a question does ask for $a$ alone, it has to give you another fact about $a$ or $b$. Put that fact in too: for $b=2$, type b=2 on a line above, and Desmos fits only $a$, here $6$.