## Quadratic Equation:

A quadratic equation is an equation in which the highest power of the variable
is two (also known as a second-degree polynomial equation). A general form of
equation is "

**ax**", where "^{2}+ bx + c = 0**a**", "**b**", and "**c**" are constants and "**x**" is the variable being solved for.
Quadratic equations are interesting because they can be graphed as a parabola.
A parabola is a curved line that, depending on the values of a, b, and c in
the equation, looks like a smile or a frown. We can graph the equation to see
what x values make the equation true, as well as where the parabola intersects
the x-axis (the horizontal line where y = 0).

### Standard form of equation:

**ax**

^{2}+ bx + c = 0#### Quadratic Equation how to solve:

To find the solutions (or roots) of the equation, we can use the

**quadratic formula**and**factor method**.#### Quadratic Formula:

**x = (-b ± √(b ^{2} - 4ac)) / 2a**

### Discriminant of a quadratic equation:

The term inside the square root, "**b**" is known as the discriminant. The equation has

^{2}- 4ac*two real solutions*if the

*discriminant is positive*.

If the

*discriminant is zero*, there is*only one real solution*to the equation.
If the discriminant is negative, there are two complex solutions to the
equation (which are conjugates of each other).

Discriminant Value | Number of Real Solutions | Nature of Solutions |
---|---|---|

Positive (b^{2} - 4ac > 0) |
Two | Real and distinct |

Zero (b^{2} - 4ac = 0) |
One | Real and repeated |

Negative (b^{2} - 4ac < 0) |
Two | Complex conjugates (non-real) |

###
Example 1: Solving equation ( 5x^{2} + 6x + 1 = 0) using quadratic formula:

To solve the equation 5x

^{2}+ 6x + 1 = 0, we can use the quadratic formula:x = [-b ± √(b

^{2}- 4ac)] / 2awhere a = 5, b = 6, and c = 1.

Substituting these values into the formula, we get:

x = [-6 ± √(6

^{2}- 4(5)(1))] / 2(5)x = [-6 ± √(36 - 20)] / 10

x = [-6 ± √16] / 10

x = [-6 ± 4] / 10

So we have two possible solutions:

x = (-6 + 4) / 10 = -1/5

x = (-6 - 4) / 10 = -1

Therefore, the solutions to the equation 5x^2 + 6x + 1 = 0 are

**x = -1/5**and**x = -1**.###
Example 2: Solving equation ( 5x^{2} + 6x + 1 = 0) using factoring:

To factor this quadratic equation, we need to find two numbers that

**multiply to give 5 and add to give 6**. These numbers are 1 and 5. So we can rewrite the equation as:5x

^{2}+ 5x + x + 1 = 0Now we can factor by grouping:

(5x

^{2}+ 5x) + (x + 1) = 05x(x + 1) + 1(x + 1) = 0

(5x + 1)(x + 1) = 0

So the solutions to the equation 5x

^{2}+ 6x + 1 = 0 are:5x + 1 = 0, which gives x = -1/5

x + 1 = 0, which gives x = -1

Therefore, the solutions to the equation 5x

^{2}+ 6x + 1 = 0 are x = -1/5 and x = -1, which is the**same answer we got using the quadratic formula**.
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