If y = x2 – 32x + 256, then what is the least possible value of y ?
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Solution
First, factor the quadratic equation: x2 – 32x + 256 = (x – 16)2. Any quantity squared is either positive or zero.To minimize the expression (x – 16)2 and the value of y, let x = 16, so that y = 0.The answer is choice (E).
If \(\frac{x^{2}-4x+4}{x^{2}+x-6}=\frac{6x-12}{x^{2}+6x+9}\), then x =
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Solution
3
To solve this question, factor everything to its simplest form: x2 – 4x + 4 factors to (x – 2) (x – 2); x2 + 6x + 9 factors to (x + 3)(x + 3); x2 + x – 6 factors to (x – 2)(x + 3); and, finally, 6x – 12 factors to 6(x – 2).Thus,\(\frac{x^{2}-4x+4}{x^{2}+x-6}=\frac{6x-12}{x^{2}+6x+9}\) factors to \(\frac{(x-2)(x-2)}{(x-2)(x+3)}=\frac{6(x-2)}{(x+3)(x+3)}\).Cancel (x – 2) from the numerator and denominator on the left-hand side of the equation to yield\(\frac{(x-2)}{(x+3)}=\frac{6(x-2)}{(x+3)(x+3)}\). Next, multiply both sides by (x + 3) to yield (x – 2) = \(\frac{(x-2)}{(x+3)}=\frac{6(x-2)}{(x+3)(x+3)}\), and divide both sides by (x – 2) to yield 1 = \(\frac{6}{(x+3)}\). Finally, multiply both sides by (x + 3) again or turn 1 into a fraction and cross-multiply; either way, you’re left with 6 = x + 3, so x = 3.
x2 – 2xy + y2 = 0 and y = 9⁄x
| Quantity A | Quantity B |
| y | 3 |
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Solution
Factor the quadratic expression to get (x – y)(x – y) = 0; x – y must equal 0, so you know that x = y.Thus, y= 9⁄y, y2 = 9, and y = 3 —eliminate choices (A) and (B)—or –3—eliminate choice (C).The answer is choice (D).
| Quantity A | Quantity B |
| (s + t)2 | s2 + t2 |
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Solution
The best approach here is to Plug In. First, try s = 2 and t = 3: Quantity A is (2 + 3)2 = 52 = 25, and Quantity B is 22 + 32 = 4 + 9 = 13. Quantity A is greater, so eliminate choices (B) and (C). Next, make s and t both 0: Now Quantity A is (0)2 = 0, and Quantity B is 02 + 02 = 0. Now the two quantities are equal, so eliminate choice (A), and you’re left with choice (D).
x ≥ 0
y ≥ 0
| Quantity A | Quantity B |
| \(\sqrt{x^{12}}-y\) | \(\sqrt{x^{12}}-y\) |
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Solution
In Quantity A,\(\sqrt{x^{12}}-y=\sqrt{(x^{6})^{2}}-y=x^{6}-y\). In Quantity B, you may recognize one of the common quadratics: (a + b)(a – b) = a2 – b2. If not, FOIL; either way, Quantity B is x6 – y.Thus, the two quantities are equal.
(–x + y)(–y + x) =
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Solution
Factor out –1 from the parentheses on the left and rearrange the expression in the parentheses on the right to get –1(x – y) (x – y) = –(x – y)2.The answer is choice (D).
| Quantity A | Quantity B |
| (141)2 – (28)2 | (141 – 28)2 |
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Solution
Don’t do the arithmetic!These are common quadratic patterns. It’s not important that x = 141 and y = 28; Quantity A is x2 – y2 = (x + y)(x – y), and Quantity B is (x – y)(x – y). Since (x – y) is a positive number, you can simply compare the remaining factors after it is removed from both quantities. Since x and y are positive, (x + y) is greater than the remaining (x – y) in Quantity B, and the answer is choice (A).
a ≠ –b
| Quantity A | Quantity B |
| \(\frac{6a^{2}+12ab+6b^{2}}{a+b}\) | 6(a + b) |
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Solution
Whenever you see exponents, think common quadratics. If you factor the 6 out of the numerator in Quantity A, you get 6(a2 + 2ab + b2), which includes a common quadratic (a + b)2.Then you can cancel (a + b) from both the numerator and the denominator; Quantity A is really just 6(a + b).The quantities are equal.
For x ≠ –2 and x ≠ –4,\(\frac{x}{x+4}+\frac{-3}{x+2}=\)
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Solution
Plug in x = 2, and the original expression turns into \(\frac{2}{6}+\frac{-3}{4}=\frac{1}{3}-\frac{3}{4}=-\frac{5}{12}\), using the Bowtie. Now plug in 2 for x in the answer choices to see which equals \(-\frac{5}{12}\). Only choice (A) doe
x2 – 49 = 0
| Quantity A | Quantity B |
| x2 – 7x | –7x + 49 |
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Solution
Remember that when a variable is squared, it yields a positive and a negative solution; hence, x2 – 49 = 0 means that x2 = 49 and x = ± 7. If x = 7, then both quantities are equal to zero. If x = –7, then both quantities are equal to 98.The answer is choice (C).