Algebra 2 is a branch of Mathematics that uses mathematical statements to describe relationships between things that vary over time. A mathematical statement describes relationship and we use letters to represent the quantity that varies as there is no fixed amount. Letters and symbols are known as variables. Mathematical statements that describe relationships are expressed using algebraic terms or equations. In an equation two things can be equal, an equation will have an equal sign. |

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Given below are the important algebra 2 topics:**

Algebraic Expression | Expression built from constants, variables, and a finite number of algebraic operations. |

Absolute value | Absolute value of a real number $x$ is the non negative value of $x$ without regard to its sign. |

Binomials | A polynomial with two terms. That is, the sum of two monomials. A polynomial equation with two terms usually joined by a plus or minus sign is called a binomial. |

Cartesian Coordinates | The usual coordinate system, originally described by Descartes, in which points are specified as distances to a set of perpendicular axes. Also called rectangular coordinates. |

Domain |
The domain of a function is the complete set of possible values of the independent variable |

Complex Numbers | A number that is expressed in the form $a + bi$, where $a$ and $b$ are real numbers and $i$ being the imaginary unit. |

Monomials | A monomial is a product of positive integer powers of a fixed set of variables together with a coefficient. |

Functions | A function relates an input to an output. Each input is exactly related to one output. |

Slope intercept form | It is used to express equation of a line, $m$ is the slope and $b$ is the $y$-intercept. |

Rational Expressions | An expression that is the ratio of two polynomials, just like a fraction. |

Root | A root of an equation is a solution of that equation. |

Tangents | Tangent is a line which touches the outer surface of the circle or circular object at exactly one point. |

Real Numbers | Real number is a value that represents a quantity along a continuous line. |

Trigonometry | Branch of mathematics that studies triangles and the relationships between the lengths of their sides and the angles between those sides. |

Variable | A symbol that represents a quantity in a mathematical expression. |

Range | The range of a function is the complete set of all possible resulting values of the dependent variable. |

Term | The parts that make up an expression that are separated by '$+$' and '$-$' signs. |

### Solved Examples

**Question 1:**Solve 5x$^{2}$ + 3x + 1 = 0

**Solution:**

In the given equation coefficients of a, b and c are 5, 3 and 1 respectively.

Formula for quadratic equation is :

x = $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Plug in the values for a, b,and c into the quadratic formula

x = $\frac{-3\pm\sqrt{3^{2}-4*5*1}}{2*5}$

x = $\frac{-3\pm\sqrt{-11}}{10}$

We see that we can have two values for x , let they be $x_{1}$ and $x_{2}$

$x_{1}$ = $\frac{-3+\sqrt{-11}}{10}$

= $\frac{-3}{10}+\frac{1}{10}\sqrt{11}$i

$x_{2}$= $\frac{-3-\sqrt{-11}}{10}$

= $\frac{-3}{10}-\frac{1}{10}\sqrt{11}$i

Therefore the values of $x_{1}$ and $x_{2}$ are $\frac{-3}{10}+\frac{1}{10}\sqrt{11}$i and

$\frac{-3}{10}-\frac{1}{10}\sqrt{11}$i respectively.

**Question 2:**Solve 4x$^{2}$ + x + 5 by using completing square method.

**Solution:**

**:**

__Step 1__As the given equation is not in standard form, divide by 4 through out the equation.

x$^{2}$ + $\frac{1}{4}$x + $\frac{5}{4}$ = 0

**:**

__Step 2__Move the constant term to the right hand side.

x$^{2}$ + $\frac{1}{4}$x = - $\frac{5}{4}$

**:**

__Step 3__Take half of the x-term coefficient and square it. Add this value to both sides.

x term coefficient = $\frac{1}{4}$

Half of the x-term coefficient = $\frac{1}{8}$

After squaring, ($\frac{1}{8})^{2}$ = $\frac{1}{64}$

Adding $\frac{1}{64}$ to both sides we get

x$^{2}$ + $\frac{1}{4}$x + $\frac{1}{64}$ = -$\frac{5}{4}$ + $\frac{1}{64}$

**:**

__Step 4__Simplifying right side

x$^{2}$ + $\frac{1}{4}$x+ $\frac{1}{64}$ = -$\frac{79}{64}$

**:**

__Step 5__Write the perfect square on the left

(x + $\frac{1}{8}$)$^{2}$ = -$\frac{79}{64}$

**:**

__Step 6__Take the square root of both sides.

x + $\frac{1}{8}$ = $\pm\sqrt{\frac{-79}{64}}$

**:**

__Step 7__Solve for x

x = - $\frac{1}{8}$ $\pm\sqrt{\frac{-79}{64}}$

or x = - $\frac{1}{8}$ $\pm$ $\frac{1}{8}\sqrt{79}$i