Radicals & Absolute Values - SAT Math

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

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

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

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

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

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

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

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

An expression of finite terms with non-negative integer exponents. Sum of terms with variables raised to whole number powers.

$(x, y) = (2, 4)$. Set equations equal: $3x - 2 = -x + 6$.

$x = 2$ or $x = 1$. Substitute $y = 3 - x$ into first equation and solve.

$x = 0$ or $x = 4$. Factor out $x$: $x(x - 4) = 0$.

$(x, y) = (3, 1)$ or $(x, y) = (-3, -1)$. Substitute $x = 3y$ into circle equation.

$(x, y) = (-\frac{2}{3}, \frac{5}{3})$. Set equations equal: $2x + 3 = -x + 1$.

$-\frac{b}{a}$. From Vieta's formulas for quadratic equations.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Standard formula for solving quadratic equations.

$(x - 2)^2$. Perfect square trinomial in factored form.

1. Substitute $x = 0$ into the equation.

$(x - 3)(x + 3)$. Difference of squares pattern: $a^2 - b^2$.

$x = 4$ or $x = -4$. Set $y = 0$ and solve $x^2 - 16 = 0$.

$(2, 0)$. First equation gives $x = 2$, then substitute into second equation.

$x = 3$ or $x = 2$. Factor as $(x - 3)(x - 2) = 0$.

1. Value where line crosses the $y$-axis.

Substitution or elimination. Both methods work by reducing the system to simpler equations.

$x = 3$ or $x = -3$. Take square root of both sides.

1. Term without any variable.

$(-3, 0)$. Perfect square $(x + 3)^2$ has vertex at $x = -3$.

$5x + 12$. Distribute 3, then combine like terms.

$2x$. Add fractions with same denominator: $\frac{3x + x}{2} = \frac{4x}{2} = 2x$.

$a + 8$. Distribute 2, then combine like terms.

$(a + b) + c = a + (b + c)$. Grouping doesn't change the sum.

$11 - 2x$. Distribute negative sign, then combine like terms.

$11x - 8$. Distribute 2, then combine like terms.

$2x + 2$. Combine like terms by adding coefficients.

$(x + y)^2$. Perfect square trinomial pattern: $a^2 + 2ab + b^2 = (a + b)^2$.

$-2x + 6$. Distribute then combine like terms: $2x + 6 - 4x = -2x + 6$.

$2x$. Distribute -2, then combine like terms.

$x + 6$. Distribute -3, then combine like terms.

$a \times 1 = a$. Multiplying by 1 doesn't change the value.

$5x - 3$. Distribute 3, then combine like terms.

$9x^2 - 12x + 4$. Expand using $(a - b)^2 = a^2 - 2ab + b^2$.

$5x - 3$. Distribute negative sign, then combine like terms.

$ab = ba$. Order doesn't matter in multiplication.

$6y - 12$. Distribute 4, then combine like terms.

$4a + 8$. Distribute 4 to both terms inside parentheses.

$a \times \frac{1}{a} = 1$ for $a \neq 0$. Multiplying by reciprocal gives 1.

$-x + 10$. Distribute 2, then combine like terms.

7x. Combine like terms by adding coefficients.

$-a + 8$. Distribute -4, then combine like terms.

$x^2 - 2x$. Distribute $x$, then combine like terms.

$a(b + c) = ab + ac$. Multiplying distributes over addition inside parentheses.

$2x + 6$. Distribute 3, then combine like terms.

$2x + 11$. Distribute 2, then add 3.

$7x - 5$. Distribute 5, then combine like terms.

$a + 0 = a$. Adding zero doesn't change the value.

$(a - b)^2$. Perfect square trinomial pattern: $a^2 - 2ab + b^2 = (a - b)^2$.

$(2x - 3)(2x + 3)$. Difference of squares: $(2x)^2 - 3^2 = (2x - 3)(2x + 3)$.

$-x + 10$. Distribute both terms, then combine like terms.

$5x$. Combine like terms by adding coefficients.

$5a + 5b$. Distribute 5 to both terms inside parentheses.

$7y + 4$. Group like terms and combine coefficients.

$7x^2$ and $-2x^2$. Like terms have the same variable and exponent.

$(x - 2)(x + 2)$. Difference of squares: $a^2 - b^2 = (a - b)(a + b)$.

$5x - 4$. Distribute and combine: $2x - 10 + 3x + 6 = 5x - 4$.

$x + 4$. Distribute negative, then combine like terms.

$x = 4$ or $x = -2$. Factor as $(x - 4)(x + 2) = 0$.

Degree 3. The highest degree among all terms in the system is 3.

$5x^3$. Term with highest degree in the polynomial.

$x = i$ or $x = -i$. Complex roots when discriminant is negative.

1. Highest power of the variable term.

1. Coefficient of the highest degree term.

1. Highest power of the variable in the polynomial.

Non-linear. The presence of $x^2 + y^2$ makes the system non-linear.

$ax^2 + bx + c = 0$. Standard form with $a \neq 0$.

$(1, 2)$ and $(2, 1)$. Both points satisfy both equations when substituted.

$(1, -4)$. Vertex form shows vertex at $(h, k)$.

$x = 1$ or $x = -1$. Set $x^2 - 1 = 0$ and factor.

$5$. Highest power determines polynomial degree.

$f(x, y) = 0, , g(x, y) = 0$. Two polynomial equations set equal to zero that must be solved simultaneously.

$y = 2$. Perfect square trinomial with double root.

$x = -3$. Perfect square trinomial $(x + 3)^2 = 0$.

$(x, y) = (3, 2)$. Add equations to get $2x = 4$, then substitute back.

$(x, y) = (1, \frac{5}{3})$. Solve by elimination or substitution method.

$x = 2$ or $x = -2$. Set $x^2 - 4 = 0$ and solve.

$x = -1$. Perfect square trinomial $(x + 1)^2 = 0$.

$ (x - 3)(x + 1) $. Factor by finding two numbers that multiply to $-3$.

$y = (x - 2)^2$. Complete the square to get vertex form.

$\frac{c}{a}$. From Vieta's formulas for quadratic equations.

$b^2 - 4ac$. Determines nature of roots for quadratic equations.

Intersection points of the graphs. Each solution represents a point where all equations are satisfied.

Graph the equations and find intersection points. Solutions occur where the curves visually cross on the coordinate plane.

$(3, 9)$. Substitute $y = 3x$ into the first equation to find $x = 3$.

Opens upwards. Positive coefficient of $x^2$ means parabola opens upward.

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

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

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

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

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

1. Using $b^2 - 4ac = 16 - 16 = 0$, indicating one repeated root.

$a^2 - b^2 = (a-b)(a+b)$. A fundamental factoring pattern for squared terms.

The coefficient is -3. The numerical factor multiplying $x^2$.

The sum of the roots is 6. By Vieta's formulas: $-\frac{b}{a} = -\frac{(-6)}{1} = 6$.

The remainder is 0. By the Remainder Theorem, substitute $x = 1$.

The leading term is $3x^4$. The term with the highest degree and its coefficient.

The zero is 5. Set the polynomial equal to zero and solve.

$x^2 - x - 6$. Use FOIL to multiply the binomials.

The degree is 4. The highest power of x determines the degree.

$(x - 2)(x + 2)$. Difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

The linear term is $2x$. The term with $x$ raised to the first power.

$ax^2 + bx + c = 0$. The general form of any quadratic equation.

Yes. Substituting: $(2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0$.

The product of the roots is 4. By Vieta's formulas: $\frac{c}{a} = \frac{4}{1} = 4$.

$-\frac{b}{a}$. By Vieta's formulas, sum of roots equals $-\frac{b}{a}$.

1. The highest power of $x$ determines the polynomial's degree.

The constant term is 1. The term with no variable, also called the constant coefficient.

The multiplicity is 2. The exponent of $x$ indicates the multiplicity of zero.

$ax^2 + bx + c = 0$. The general form where $a \neq 0$ defines a quadratic equation.

The degree is 5. The highest power of the variable determines the degree.

$a_n x^n + a_{n-1} x^{n-1} + \text{...} + a_1 x + a_0 = 0$. Standard polynomial equation with terms arranged by decreasing powers.

The sum is $-\frac{b}{a}$. Vieta's formula relating coefficients to root sum.

The quadratic term is $-5x^2$. The term with $x$ raised to the second power.

Example: $x^3 + 2x^2 + x + 1$. Any polynomial where the highest power is 3.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Derived by completing the square on the standard form.

It is called a monomial. A polynomial with exactly one term.

$\frac{5}{2}$. By Vieta's formulas, product of roots equals $\frac{c}{a} = \frac{5}{2}$.

It is a quartic polynomial. A degree 4 polynomial is called quartic.

There are 3 zeros. A cubic polynomial can have at most 3 real zeros.

The y-intercept is 7. The constant term gives the y-intercept.

The leading coefficient is 5. The coefficient of the highest degree term.

The GCF is $3x^2$. Factor out the common term $3x^2$.

1. The coefficient of the highest degree term is the leading coefficient.

The multiplicity is 2. The exponent of $(x-2)$ indicates its multiplicity.

$x - 3$. Factor and cancel: $\frac{(x-3)(x+3)}{(x+3)} = x-3$.

$(x - 3)^2$. This is a perfect square trinomial: $a^2 - 2ab + b^2 = (a-b)^2$.

$(x - 3)(x + 3)$. This is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

The product is $\frac{c}{a}$. Vieta's formula relating coefficients to root product.

$x = -2$. Perfect square trinomial $(x + 2)^2 = 0$.

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

1. Absolute value removes the negative sign.

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

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

1. Zero has no distance from itself.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1. Absolute value makes any number non-negative.

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

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

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

1. Absolute value makes negative numbers positive.

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

$(x + 2)(x + 3)$. Find two numbers that multiply to 6 and add to 5.

If $b^2 - 4ac > 0$, 2 real roots; $=0$, 1 real root; $<0$, no real roots. The discriminant determines whether roots are real or complex.

Intercept: $(0, 1)$. Set $x = 0$ to find where the parabola crosses the $y$-axis.

Vertex: $(2, 1)$. Use $x = -\frac{b}{2a} = 2$, then $y = -(4) + 8 - 3 = 1$.

Minimum: $0$. This is $(x-3)^2$, so minimum occurs at vertex $(3,0)$.

$x = 2, x = 3$. Factor as $(x-2)(x-3) = 0$ or use the quadratic formula.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Derived from completing the square on the general quadratic equation.

Vertex: $(2, -1)$. Use $x = \frac{4}{2} = 2$, then $y = 4 - 8 + 3 = -1$.

$ax^2 + bx + c = 0$. General form where $a \neq 0$ determines parabola shape.

$b^2 - 4ac$. The expression under the square root in the quadratic formula.

$(x - 3)(x + 3)$. Difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

$b^2 = 4ac$. When discriminant equals zero, the quadratic is a perfect square.

$x = 2, x = -4$. Factor as $(x-2)(x+4) = 0$ or use the quadratic formula.

$x = \frac{-b}{2a}$. The $x$-coordinate of the vertex, found by calculus or completing the square.

$x = -2, x = 2$. Set $y = 0$: $x^2 - 4 = 0$, so $x^2 = 4$.

If $ab = 0$, then $a = 0$ or $b = 0$. If a product equals zero, at least one factor must be zero.

$y = a(x-h)^2 + k$. Shows vertex $(h,k)$ and vertical stretch/compression factor $a$.

$ax^2 + bx + c = 0$. The general form where $a \neq 0$ defines a parabola.

Narrower parabola. Larger $|a|$ values compress the parabola horizontally.

$y = a(x-h)^2 + k$. Shows vertex $ (h,k) $ and vertical shift directly.

$(x + 4)^2$. Perfect square trinomial with $a = x$, $b = 4$.

$x^2 + 4x + 4$. Apply the perfect square formula $(a+b)^2 = a^2 + 2ab + b^2$.

$b^2 - 4ac$. Expression under the square root that determines root types.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Solves any quadratic by substituting coefficients $a$, $b$, and $c$.

$\frac{c}{a}$. By Vieta's formulas relating coefficients to roots.

$y = (x-2)^2$. This is $(x-2)^2$ expanded, so vertex form shows vertex $(2,0)$.

$x = \frac{-b}{2a}$. Derived from completing the square or using calculus.

$y = 2x^2 - 12x + 22$. Expand $(x-3)^2 = x^2 - 6x + 9$, then distribute and combine.