ACT and SAT Math Tips: Classic Quadratic Forms

I’ve said it before and I’ll say it again: the writers of standardized tests assess the same concepts, over and over and over again.  Test experts are able to become test experts for that very reason—there is a finite amount of material one needs to learn in order to master the ACT and the SAT. If you study a few dozen tests and find these predictable patterns, you could be an expert, too.

Luckily, I’m here to save you from that mind-numbing task. And trust me—it’s mind-numbing, especially when we get into some of the concepts from Algebra II, such as classic quadratic forms.

         1.            (x + y)(x – y)  =  x² – y²

                         Examples:

                        (t + 5)(t – 5)   =  t² – 5²   =   t² – 25

                        (3a + b)(3a – b)  =  (3a)² – b²  =  9a² – b²

                        y² – 64  =  (y + 8)(y – 8)

                        36 – n²  =  (6 + n)(6 – n)

        2.            (x + y)²  =  x² + 2xy + y²

                        Examples:

                        (t + 5)²  =  t² + 2(t)(5) + 5²  =  t² + 10t + 25

                        (3a + b)²  =  (3a)² + 2(3a)(b) + b²  =  9a² + 6ab + b²

                        (6 + n)²  = 6² + 2(6)(n) + n²  = 36 + 12n + n²

        3.            (x – y)²  =  x² – 2xy + y²

                        Examples:

                        (t – 5)²  =  t² – 2(t)(5) + 5²  =  t² – 10t + 25

                        (3a – b)²  =  (3a)² – 2(3a)(b) + b²  =  9a² – 6ab + b²

                        (6 – n)²   = 6² – 2(6)(n) + n²  = 36 – 12n + n²

Now let’s look at how memorizing these forms can help you quickly and confidently solve an ACT or SAT question.

1.     If r – p = 7, what is the value of r² – 2rp + p²?

        (A)     7

        (B)     14

        (C)     28

        (D)     35

        (E)     49

A student who doesn’t know the classic ACT and SAT quadratic equations will start to (unsuccessfully) use substitution:

r – p = 7

r = 7 + p

They will plug 7 + p into the second expression for r, even though the second expression is just an expression—not an equation!

r² – 2rp + p²

(7 + p)² – 2(7 + p)p + p²

(7 + p) (7 + p) – 2(7 + p)p + p²

49 + 14p + 2p² – 2(7p + p²) + p²

49 + 14p + 2p² – 14p – 2p² + p²

49 + 2p²– 2p² + p²

49 + p²

Oh, wait! It’s just an expression. It doesn’t equal anything. This won’t work!

A lot of wasted time for the average student.

But for you, the reader of great test prep blogs, this question is solved in about 5 seconds. You look at r² – 2rp + p² and immediately recognize the third classic quadratic form:

(x – y)²  =  x² – 2xy + y²

So (r – p)² =  r² – 2rp + p²  

And since r – p = 7, then (r – p)² =  7² = 49.

Voila.

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