Radicals & Absolute Values
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SAT Math › Radicals & Absolute Values
Simplify the expression: $\frac{12}{\sqrt{3}}$.
12$\sqrt{3}$
$\frac{12\sqrt{3}$}{3}
3$\sqrt{3}$
4$\sqrt{3}$
Explanation
This question requires rationalizing the denominator of the fraction 12/√3 to eliminate the radical from the denominator. To rationalize, we multiply both numerator and denominator by √3: (12/√3) × (√3/√3) = 12√3/3. Finally, we simplify the fraction by dividing: 12√3/3 = 4√3.
What is the solution to the equation $\sqrt{x + 1} = 4$?
3
15
16
17
Explanation
This question asks us to solve the radical equation √(x + 1) = 4 by isolating the variable under the radical. To eliminate the square root, we square both sides: (√(x + 1))² = 4², which gives us x + 1 = 16. Solving for x: x = 16 - 1 = 15.
Which expression is equivalent to $\sqrt{18} + \sqrt{8}$?
$\sqrt{26}$
3$\sqrt{2}$ + 2$\sqrt{2}$
5$\sqrt{2}$
$\sqrt{18 + 8}$
Explanation
This question asks us to find an equivalent expression for √18 + √8 by simplifying each radical separately and then combining like terms. First, simplify √18 = 3√2 and √8 = 2√2. Now we can add the like terms: 3√2 + 2√2 = 5√2.
Simplify the expression: $\sqrt{72}$.
8.5
$\sqrt{36}$ + $\sqrt{36}$
12
6$\sqrt{2}$
Explanation
This question asks us to simplify the radical expression √72 by factoring out perfect squares. First, we find the prime factorization: 72 = 36 × 2. Since 36 is a perfect square, we can extract it from under the radical: √72 = √(36 × 2) = √36 × √2 = 6√2.
$$ \sqrt{x + 14} = x + 2 $$
What is the solution to the given equation?
-5
2
4
5
Explanation
1. Square both sides: $x + 14 = (x + 2)^2$.
2. Expand: $x + 14 = x^2 + 4x + 4$.
3. Set to zero: $x^2 + 3x - 10 = 0$.
4. Factor: $(x + 5)(x - 2) = 0$.
5. Potential solutions: $x = -5, x = 2$.
6. Check for extraneous solutions:
- If $x = -5$: $\sqrt{9} = -3$ (False, principal root must be positive).
- If $x = 2$: $\sqrt{16} = 4$ (True).
Only $x=2$ is valid.
SAT Strategy: Backsolve
Plug the answer choices into the equation.
A) $\sqrt{-5+14} = -5+2 \rightarrow 3 = -3$ (False).
B) $\sqrt{2+14} = 2+2 \rightarrow 4 = 4$ (True).
Which of the following describes the set of all real numbers that are 8 units away from -2?
$|x+2|=-8$
$|x-2|=8$
$|x+2|=8$
$|x-2|=-8$
Explanation
Even if you're unsure of where to start on this problem, you should have a head start. The problem is testing absolute values, and you should know that the result of any absolute value is always nonnegative ≥0. So the answer choices that include an absolute value equaling a negative number must be incorrect: that just cannot be possible.
To test the remaining choices, consider that the numbers that are exactly eight units away from -2 are -2+8 = 6, and -2-8 = -10. When you plug these numbers in for x in the answer choices, only one is valid:
$|x+2|=8$
For $x=6$, $|6+2|=|8|=8$
For $x=-10$, $|-10+2|=|-8|=8$
Therefore this absolute value satisfies the given situation and is correct.