Radicals & Absolute Values - SAT Math

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

Vertex: $(1, 3)$. In vertex form $y = a(x-h)^2 + k$, vertex is $(h,k)$.

Vertex: $(-2, 0)$. This is $(x+2)^2$, so vertex is at $x = -2$, $y = 0$.

The quadratic formula. Works for any quadratic, unlike factoring or square roots.

$x = -1$. Use $x = -\frac{b}{2a} = -\frac{6}{2(3)} = -1$.

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

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

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

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

$x = 2, x = -2$. Divide by 3: $x^2 - 4 = 0$, so $x = \pm 2$.

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

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

Discriminant: $0$. Calculate $b^2 - 4ac = 16 - 16 = 0$.

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

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

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

$y = (x+3)^2 - 1$. Complete the square: $(x+3)^2 = x^2 + 6x + 9$, so subtract 1.

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

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

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

A downward-opening parabola. Negative coefficient of $x^2$ makes the parabola open downward.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1. Coefficient of the highest degree term.

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

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

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

1. Term without any variable.

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

1. Highest power determines polynomial degree.

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

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

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

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

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

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

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

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

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

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

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

1. Highest power of the variable term.

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

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

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

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

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

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

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

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

1. Highest power of the variable in the polynomial.

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

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

$|\text{a} + \text{b}|$. 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.

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

1. Zero has no distance from itself.

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

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

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