Radicals & Absolute Values - SAT Math

1. Since $9 \times 9 = 81$.

1. Since $3^3 = 27$, the cube root is 3.

1. Calculate $5 - 9 = -4$, then take absolute value to get 4.

Not a real number. Square roots of negative numbers aren't real.

$x = 12$ or $x = -12$. Absolute value equation has two solutions: $x = \pm 12$.

0.5. Since $0.5 \times 0.5 = 0.25$.

$x = 4$ or $x = -1$. Distance equals 5, so $2x - 3 = \pm 5$.

1. Absolute value makes negative numbers positive.

$2\text{√}5$. Factor out perfect squares: $\sqrt{20} = \sqrt{4 \times 5}$.

$3\text{√}3$. Factor out perfect squares: $\sqrt{27} = \sqrt{9 \times 3}$.

$\sqrt{9} = 3$. Simplify by dividing: $\frac{\sqrt{18}}{\sqrt{2}} = \sqrt{\frac{18}{2}} = \sqrt{9}$.

1. Absolute value makes any number non-negative.

$x = 8$ or $x = -2$. Distance from 3 equals 5, so $x - 3 = 5$ or $x - 3 = -5$.

$10\text{√}2$. Factor out perfect squares: $\sqrt{200} = \sqrt{100 \times 2}$.

$5\text{√}2$. Factor out perfect squares: $\sqrt{50} = \sqrt{25 \times 2}$.

$|\text{a} \times \text{b}| = |\text{a}| \times |\text{b}|$. The absolute value of a product equals the product of absolute values.

$|\text{a} + \text{b}|$. Triangle inequality shows they're not always equal.

1. Absolute value gives the distance from zero, always positive.

$\sqrt{25} = 5$. This simplifies to $\sqrt{25} = 5$ using the quotient property of radicals.

$x = 7$ or $x = 1$. Distance from 4 equals 3, so $x - 4 = \pm 3$.

1. Since $7 \times 7 = 49$.

$4\text{√}2$. Factor out perfect squares: $\sqrt{32} = \sqrt{16 \times 2}$.

1. The square of a square root equals the radicand.

1. Calculate $3 - 7 = -4$, then take absolute value.

$3\text{√}3 - \text{√}3 = 2\text{√}3$. Simplify radicals first: $\sqrt{12} = 2\sqrt{3}$.

$6\text{√}2$. Factor out perfect squares: $\sqrt{72} = \sqrt{36 \times 2}$.

1. Zero has no distance from itself.

1. Since $8^2 = 64$, the principal square root is 8.

$x = 3$ or $x = -\frac{13}{3}$. Distance equals 11, so $3x + 2 = \pm 11$.

1. Absolute value removes the negative sign.

$x = 25$. Square both sides to eliminate the radical.

$6\sqrt{3}$. Simplify to $4\sqrt{3} + 2\sqrt{3} = 6\sqrt{3}$.

$|\text{a} \times \text{b}|$. Absolute values multiply to give the absolute value of the product.

1. Since $1 \times 1 = 1$.

$2\text{√}2 + 3\text{√}2 = 5\text{√}2$. Simplify each radical first: $\sqrt{8} = 2\sqrt{2}$, $\sqrt{18} = 3\sqrt{2}$.

1. Since $8 \times 8 = 64$.

$6\text{sqrt}(2)$. Factor out perfect squares: $\sqrt{36 \times 2} = 6\sqrt{2}$.

$x = 3$ or $x = -13$. Distance from -5 equals 8, so $x + 5 = \pm 8$.

$|\text{a} \times \text{b}|$. Absolute values multiply to give the absolute value of the product.

1. Zero has no distance from itself.

Not a real number. Square roots of negative numbers aren't real.

$4\text{√}2$. Factor out perfect squares: $\sqrt{32} = \sqrt{16 \times 2}$.

$|\text{a} + \text{b}|$. Triangle inequality shows they're not always equal.

1. Absolute value removes the negative sign.

$x = 4$ or $x = -1$. Distance equals 5, so $2x - 3 = \pm 5$.

1. Since $8 \times 8 = 64$.

$x = 3$ or $x = -13$. Distance from -5 equals 8, so $x + 5 = \pm 8$.

0.5. Since $0.5 \times 0.5 = 0.25$.

1. Calculate $3 - 7 = -4$, then take absolute value.

1. The square of a square root equals the radicand.

$2\text{√}2 + 3\text{√}2 = 5\text{√}2$. Simplify each radical first: $\sqrt{8} = 2\sqrt{2}$, $\sqrt{18} = 3\sqrt{2}$.

$3\text{√}3 - \text{√}3 = 2\text{√}3$. Simplify radicals first: $\sqrt{12} = 2\sqrt{3}$.

$6\text{√}2$. Factor out perfect squares: $\sqrt{72} = \sqrt{36 \times 2}$.

$\text{√}9 = 3$. Divide radicals: $\frac{\sqrt{45}}{\sqrt{5}} = \sqrt{\frac{45}{5}} = \sqrt{9}$.

$2\text{√}5$. Factor out perfect squares: $\sqrt{20} = \sqrt{4 \times 5}$.

1. Since $7 \times 7 = 49$.

$x = 7$ or $x = 1$. Distance from 4 equals 3, so $x - 4 = \pm 3$.

1. Absolute value makes negative numbers positive.

$3\text{√}3$. Factor out perfect squares: $\sqrt{27} = \sqrt{9 \times 3}$.

$10\text{√}2$. Factor out perfect squares: $\sqrt{200} = \sqrt{100 \times 2}$.

$x = 12$ or $x = -12$. Absolute value equation has two solutions: $x = \pm 12$.

1. Since $9 \times 9 = 81$.

$x = 3$ or $x = -\frac{13}{3}$. Distance equals 11, so $3x + 2 = \pm 11$.

1. Since $1 \times 1 = 1$.

$5\text{√}2$. Factor out perfect squares: $\sqrt{50} = \sqrt{25 \times 2}$.

$\text{√}9 = 3$. Simplify by dividing: $\frac{\sqrt{18}}{\sqrt{2}} = \sqrt{\frac{18}{2}} = \sqrt{9}$.

1. Absolute value makes any number non-negative.

$x = 25$. Square both sides to eliminate the radical.