Prove a comparison or find a counterexample

A real solution can impose a hidden bound

A fixed comparison must hold for every value the conditions allow. Read all the conditions together: a parameter in an equation is not necessarily free to take any value.

Suppose x and a are real, $x^2-6x+a=0$ , and $x\ne3$ . Completing the square gives $(x-3)^2=9-a$ . The left side is strictly positive because x differs from 3. Therefore $9>a$ , so Quantity A is greater for every allowed solution.

If the restriction $x\ne3$ were removed, $x=3$ and $a=9$ would also be allowed. The bound would become $a\le9$ , and equality would need its own check.

Quantity A

9

Quantity B

a

Given

x^2-6x+a=0

with

x\ne3

, compare 9 and a.

A bound does not fix every comparison

Keep x and a real and $x^2-6x+a=0$ , but allow $x=3$ too. Now compare a with 8. The bound $a\le9$ alone does not settle this comparison.

For $x=3$ , the equation gives $a=9>8$ . For $x=\frac72$ , it gives $a=\frac{35}{4}>8$ . Both examples make A greater than B. They show what can happen, not what must always happen.

For $x=5$ , the same equation gives $a=5<8$ . This valid case reverses the comparison. The relationship cannot be determined. Two agreeing examples never remove the need for a proof or a counterexample.

Quantity A

a

Quantity B

8

Given

x^2-6x+a=0

for real x and a, compare a with 8.

Check when an absolute-value bound becomes equal

Let x and y be real, with $(x-3)(y-3)\ne0$ . Compare the sum of the two separate absolute values with the absolute value of their sum.

Set $u=x-3$ and $v=y-3$ . Both shifted values are nonzero, but the condition does not tell you their signs. Compare an opposite-sign pair with a same-sign pair. Does one relationship hold in both cases?

Quantity A

\left|x - 3\right| + \left|y - 3\right|

Quantity B

\left|x + y - 6\right|

Given

(x-3)(y-3)\ne0

, compare the separate magnitudes with the magnitude of their sum.

Try it

If x and y are real and $(x-3)(y-3)\ne0$ , compare $A=\lvert x-3\rvert+\lvert y-3\rvert$ with $B=\lvert x+y-6\rvert$ .

Quantity A is always greater
The relationship cannot be determined

Show the answer

The relationship cannot be determined

At $x=2,y=5$ , $A=3,B=1$ . At $x=4,y=5$ , $A=B=3$ . Both pairs satisfy the nonzero-product condition. A is always at least B. Greater in one case and equal in another means the relationship cannot be determined.

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Practice Compare quantities: Variables

Prove a comparison or find a counterexample

A real solution can impose a hidden bound

Lesson help

About this lesson

A real solution can impose a hidden bound

A bound does not fix every comparison

Check when an absolute-value bound becomes equal

Try it

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