Question

1. which equation has no solution?

a) $\sqrt{x} = 4$ b) $-\sqrt{x} = -4$ c) $\sqrt{x} = -4$

1. classify each number as rational or irrational.

$9.6\overline{8}$ $2.010010001 \dots$ $\sqrt{64}$ $-\frac{51}{5}$ $\sqrt{6}$

1. determine which two integers the following square root is between.

$\sqrt{29}$

1. solve for x?

$\sqrt{x} = 49$ $\sqrt{x} + 4 = 29$

1. what is the side length, s, of the square?

$a = 81\\ m^2$
real numbers
rational or irrational
example
natural numbers
whole numbers
integers
rational
irrational
rational
irrational

Explanation:

Step1: Analyze square root domain

The principal square root $\sqrt{x}$ is defined for $x\geq0$, and $\sqrt{x}\geq0$ for all valid $x$.

For 1a:

Solve $\sqrt{x}=4$
Square both sides: $x=4^2=16$, which is valid.

For 1b:

Solve $-\sqrt{x}=-4$
Multiply by -1: $\sqrt{x}=4$, so $x=16$, valid.

For 1c:

$\sqrt{x}=-4$: $\sqrt{x}$ cannot be negative, no solution.

Step2: Classify rational/irrational

• Rational: Terminating/repeating decimals, integers, fractions, perfect square roots.

$9.6\overline{8}$ (repeating decimal), $\sqrt{64}=8$ (integer), $-\frac{51}{5}$ (fraction)

• Irrational: Non-repeating/non-terminating decimals, non-perfect square roots.

$2.010010001...$ (non-repeating), $\sqrt{6}$ (non-perfect square)

Step3: Find bounds for $\sqrt{29}$

Find integers $n,n+1$ where $n^2<29<(n+1)^2$
$5^2=25$, $6^2=36$, so $25<29<36$

Step4: Solve for $x$

First equation $\sqrt{x}=49$:

Square both sides: $x=49^2=2401$

Second equation $\sqrt{x}+4=29$:

Subtract 4: $\sqrt{x}=29-4=25$
Square both sides: $x=25^2=625$

Step5: Find square side length

Area of square $A=s^2$, so $s=\sqrt{A}$
$s=\sqrt{81}=9$

Answer:

1. c) $\sqrt{x} = -4$
2. Rational: $9.6\overline{8}$, $\sqrt{64}$, $-\frac{51}{5}$; Irrational: $2.010010001...$, $\sqrt{6}$
3. $\sqrt{29}$ is between 5 and 6
4. $\sqrt{x}=49$: $x=2401$; $\sqrt{x}+4=29$: $x=625$
5. $s=9$ m