### Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

-21*q+11-(6*q^2)=0

## Step by step solution :

## Step 1 :

#### Equation at the end of step 1 :

` (11 - 21q) - (2•3q`^{2}) = 0

## Step 2 :

## Step 3 :

#### Pulling out like terms :

3.1 Pull out like factors:

-6q^{2} - 21q + 11=-1•(6q^{2} + 21q - 11)

#### Trying to factor by splitting the middle term

3.2Factoring 6q^{2} + 21q - 11

The first term is, 6q^{2} its coefficient is 6.

The middle term is, +21q its coefficient is 21.

The last term, "the constant", is -11

Step-1 : Multiply the coefficient of the first term by the constant 6•-11=-66

Step-2 : Find two factors of -66 whose sum equals the coefficient of the middle term, which is 21.

-66 | + | 1 | = | -65 | ||

-33 | + | 2 | = | -31 | ||

-22 | + | 3 | = | -19 | ||

-11 | + | 6 | = | -5 | ||

-6 | + | 11 | = | 5 | ||

-3 | + | 22 | = | 19 | ||

-2 | + | 33 | = | 31 | ||

-1 | + | 66 | = | 65 |

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

#### Equation at the end of step 3 :

` -6q`^{2} - 21q + 11 = 0

## Step 4 :

#### Parabola, Finding the Vertex:

4.1Find the Vertex ofy = -6q^{2}-21q+11Parabolas have a highest or a lowest point called the Vertex.Our parabola opens down and accordingly has a highest point (AKA absolute maximum). We know this even before plotting "y" because the coefficient of the first term,-6, is negative (smaller than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x-intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Aq^{2}+Bq+C,the q-coordinate of the vertex is given by -B/(2A). In our case the q coordinate is -1.7500Plugging into the parabola formula -1.7500 for q we can calculate the y-coordinate:

y = -6.0 * -1.75 * -1.75 - 21.0 * -1.75 + 11.0

or y = 29.375

#### Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = -6q^{2}-21q+11

Axis of Symmetry (dashed) {q}={-1.75}

Vertex at {q,y} = {-1.75,29.38}

q-Intercepts (Roots) :

Root 1 at {q,y} = { 0.46, 0.00}

Root 2 at {q,y} = {-3.96, 0.00}

#### Solve Quadratic Equation by Completing The Square

4.2Solving-6q^{2}-21q+11 = 0 by Completing The Square.Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:

6q^{2}+21q-11 = 0Divide both sides of the equation by 6 to have 1 as the coefficient of the first term :

q^{2}+(7/2)q-(11/6) = 0

Add 11/6 to both side of the equation :

q^{2}+(7/2)q = 11/6

Now the clever bit: Take the coefficient of q, which is 7/2, divide by two, giving 7/4, and finally square it giving 49/16

Add 49/16 to both sides of the equation :

On the right hand side we have:

11/6+49/16The common denominator of the two fractions is 48Adding (88/48)+(147/48) gives 235/48

So adding to both sides we finally get:

q^{2}+(7/2)q+(49/16) = 235/48

Adding 49/16 has completed the left hand side into a perfect square :

q^{2}+(7/2)q+(49/16)=

(q+(7/4))•(q+(7/4))=

(q+(7/4))^{2}

Things which are equal to the same thing are also equal to one another. Since

q^{2}+(7/2)q+(49/16) = 235/48 and

q^{2}+(7/2)q+(49/16) = (q+(7/4))^{2}

then, according to the law of transitivity,

(q+(7/4))^{2} = 235/48

We'll refer to this Equation as Eq. #4.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(q+(7/4))^{2} is

(q+(7/4))^{2/2}=

(q+(7/4))^{1}=

q+(7/4)

Now, applying the Square Root Principle to Eq.#4.2.1 we get:

q+(7/4)= √ 235/48

Subtract 7/4 from both sides to obtain:

q = -7/4 + √ 235/48

Since a square root has two values, one positive and the other negative

q^{2} + (7/2)q - (11/6) = 0

has two solutions:

q = -7/4 + √ 235/48

or

q = -7/4 - √ 235/48

Note that √ 235/48 can be written as

√235 / √48

### Solve Quadratic Equation using the Quadratic Formula

4.3Solving-6q^{2}-21q+11 = 0 by the Quadratic Formula.According to the Quadratic Formula,q, the solution forAq^{2}+Bq+C= 0 , where A, B and C are numbers, often called coefficients, is given by :

-B± √B^{2}-4AC

q = ————————

2A In our case,A= -6

B=-21

C= 11 Accordingly,B^{2}-4AC=

441 - (-264) =

705Applying the quadratic formula :

21 ± √ 705

q=——————

-12 √ 705 , rounded to 4 decimal digits, is 26.5518

So now we are looking at:

q=(21± 26.552 )/-12

Two real solutions:

q =(21+√705)/-12=7/-4-1/12√ 705 = -3.963

or:

q =(21-√705)/-12=7/-4+1/12√ 705 = 0.463

## Two solutions were found :

- q =(21-√705)/-12=7/-4+1/12√ 705 = 0.463
- q =(21+√705)/-12=7/-4-1/12√ 705 = -3.963