## The Quadratic Formula

The technique of completing the square deserve to be applied to any type of quadratic polynomial.

You simply rewrite* ax*2+*bx*+*c* = *a*(*x*2+

*x*)+

*c*

From it us can obtain the following result:

The roots of *ax*2+*bx*+*c* are given by

*)*

**Quadratic Formula**The quantity *b*2−4*ac** *is dubbed the * discriminant* that the polynomial.

*b*2−4

*ac*the equation has actually no actual number solutions, yet it walk have complicated solutions. If

*b*2−4

*ac*= 0 the equation has actually a repeated actual number root. If

*b*2−4

*ac*> 0 the equation has two distinctive real number roots.

### Example

Study some of these examples:

You are watching: Square root of b^2-4ac

** **

uncover the root of *x*2 + *x* + = 0

x = ± sqrt( 2 − 4× × ) 2× x = | ± sqrt( | ) | |

x = | , |

### Example

*x*2 + *x* +

**b2 - 4ac** =

**roots**

*x*1 = **, x2 = **

### Exercise

Now shot some of this exercises:

** **

The root of *x*2 + *x* + are:

Working area:

** **

## Parabola Vertex

Note that if the roots of a quadratic equation*ax*2+*bx*+*c* space real and distinct, climate the peak of the parabola offered by the polynomial is positioned where

### Example

Study a few of this examples:

** **

Locating the crest of the parabola offered by *x*2 + *x* + :

The *x*-coordinate is

*x*=

= | |

2× |

Substituting this value of x right into the given equation we find:

the *y*-coordinate is ( )2 + ( ) + =

Hence the vertex is ( , )

### Exercise

Now try some of these exercises. Provide your answer rounded come 2 decimal places:

** **

locate the crest of the parabola given by *x*2 + *x* + :

Working area:

The crest is ( , )

If the roots of a quadratic equation *ax*2+*bx*+*c* are α and also β, then we can write *ax*2+*bx*+*c* = *a*(*x*−α)(*x*−β)

completing the Square | Quadratic Polynomials index | Quadratic functions Factoriser >>

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