Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : x^2-5*x-3-(6)=0

Step by step solution :

Step 1 :

Trying to factor by splitting the middle term

1.1Factoring x2-5x-9 The first term is, x2 its coefficient is 1.The middle term is, -5x its coefficient is -5.The last term, “the constant”, is -9Step-1 : Multiply the coefficient of the first term by the constant 1•-9=-9Step-2 : Find two factors of -9 whose sum equals the coefficient of the middle term, which is -5.

Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored

Equation at the end of step 1 :

x2 – 5x – 9 = 0

Step 2 :

Parabola, Finding the Vertex:2.1Find the Vertex ofy = x2-5x-9Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up and accordingly has a lowest point (AKA absolute minimum).We know this even before plotting “y” because the coefficient of the first term,1, is positive (greater 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,Ax2+Bx+C,the x-coordinate of the vertex is given by -B/(2A). In our case the x coordinate is 2.5000Plugging into the parabola formula 2.5000 for x we can calculate the y-coordinate:y = 1.0 * 2.50 * 2.50 – 5.0 * 2.50 – 9.0 or y = -15.250Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-5x-9 Axis of Symmetry (dashed) {x}={ 2.50} Vertex at {x,y} = { 2.50,-15.25} x-Intercepts (Roots) : Root 1 at {x,y} = {-1.41, 0.00} Root 2 at {x,y} = { 6.41, 0.00}

Solve Quadratic Equation by Completing The Square

2.2Solvingx2-5x-9 = 0 by Completing The Square.Add 9 to both side of the equation : x2-5x = 9Now the clever bit: Take the coefficient of x, which is 5, divide by two, giving 5/2, and finally square it giving 25/4Add 25/4 to both sides of the equation :On the right hand side we have:9+25/4or, (9/1)+(25/4)The common denominator of the two fractions is 4Adding (36/4)+(25/4) gives 61/4So adding to both sides we finally get:x2-5x+(25/4) = 61/4Adding 25/4 has completed the left hand side into a perfect square :x2-5x+(25/4)=(x-(5/2))•(x-(5/2))=(x-(5/2))2 Things which are equal to the same thing are also equal to one another.

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Sincex2-5x+(25/4) = 61/4 andx2-5x+(25/4) = (x-(5/2))2 then, according to the law of transitivity,(x-(5/2))2 = 61/4We”ll refer to this Equation as Eq. #2.2.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of(x-(5/2))2 is(x-(5/2))2/2=(x-(5/2))1=x-(5/2)Now, applying the Square Root Principle to Eq.#2.2.1 we get:x-(5/2)= √ 61/4 Add 5/2 to both sides to obtain:x = 5/2 + √ 61/4 Since a square root has two values, one positive and the other negativex2 – 5x – 9 = 0has two solutions:x = 5/2 + √ 61/4 orx = 5/2 – √ 61/4 Note that √ 61/4 can be written as√61 / √4which is √61 / 2

Solve Quadratic Equation using the Quadratic Formula

2.3Solvingx2-5x-9 = 0 by the Quadratic Formula.According to the Quadratic Formula,x, the solution forAx2+Bx+C= 0 , where A, B and C are numbers, often called coefficients, is given by :-B± √B2-4ACx = ————————2A In our case,A= 1B= -5C= -9 Accordingly,B2-4AC=25 – (-36) = 61Applying the quadratic formula : 5 ± √ 61 x=—————2 √ 61 , rounded to 4 decimal digits, is 7.8102So now we are looking at:x=(5± 7.810 )/2Two real solutions:x =(5+√61)/2= 6.405 or:x =(5-√61)/2=-1.405