## Step 1 :

Trying to variable by dividing the center term1.1Factoring x2-6x+13 The very first term is, x2 its coefficient is 1.The center term is, -6x that is coefficient is -6.The critical term, "the constant", is +13Step-1 : multiply the coefficient of the first term by the constant 1•13=13Step-2 : find two components of 13 whose sum equals the coefficient of the center term, which is -6.

 -13 + -1 = -14 -1 + -13 = -14 1 + 13 = 14 13 + 1 = 14

Observation : No two such determinants can be uncovered !! Conclusion : Trinomial have the right to not it is in factored

Equation at the end of step 1 :

x2 - 6x + 13 = 0

## Step 2 :

Parabola, finding the Vertex:2.1Find the peak ofy = x2-6x+13Parabolas have a highest or a lowest suggest called the Vertex.Our parabola opens up and as necessary has a lowest point (AKA pure minimum).We know this even before plotting "y" because the coefficient the the very first term,1, is optimistic (greater than zero).Each parabola has actually a vertical heat of symmetry that passes v its vertex. Thus symmetry, the heat of symmetry would, because that example, pass v the midpoint that the two x-intercepts (roots or solutions) the the parabola. The is, if the parabola has actually indeed two genuine solutions.Parabolas have the right to model numerous real life situations, such as the height above ground, of things thrown upward, ~ some duration of time. The peak of the parabola can administer us through information, such as the maximum elevation that object, thrown upwards, deserve to reach. Hence we want to be able to find the coordinates of the vertex.For any type of parabola,Ax2+Bx+C,the x-coordinate that the vertex is given by -B/(2A). In our situation the x coordinate is 3.0000Plugging into the parabola formula 3.0000 because that x we can calculate the y-coordinate:y = 1.0 * 3.00 * 3.00 - 6.0 * 3.00 + 13.0 or y = 4.000

Parabola, Graphing Vertex and also X-Intercepts :

Root plot for : y = x2-6x+13 Axis of symmetry (dashed) x= 3.00 Vertex at x,y = 3.00, 4.00 function has no actual roots

Solve Quadratic Equation by completing The Square

2.2Solvingx2-6x+13 = 0 by perfect The Square.Subtract 13 from both next of the equation :x2-6x = -13Now the clever bit: take it the coefficient the x, which is 6, division by two, giving 3, and also finally square it providing 9Add 9 to both political parties of the equation :On the best hand side us have:-13+9or, (-13/1)+(9/1)The typical denominator of the two fractions is 1Adding (-13/1)+(9/1) gives -4/1So including to both sides we ultimately get:x2-6x+9 = -4Adding 9 has completed the left hand side right into a perfect square :x2-6x+9=(x-3)•(x-3)=(x-3)2 points which room equal come the very same thing are also equal come one another. Sincex2-6x+9 = -4 andx2-6x+9 = (x-3)2 then, according to the regulation of transitivity,(x-3)2 = -4We"ll refer to this Equation together Eq. #2.2.1 The Square root Principle says that once two things room equal, their square roots room equal.Note the the square root of(x-3)2 is(x-3)2/2=(x-3)1=x-3Now, using the Square root Principle to Eq.#2.2.1 us get:x-3= √ -4 add 3 to both sides to obtain:x = 3 + √ -4 In Math,iis referred to as the imagine unit. The satisfies i2=-1. Both i and also -i are the square root of -1Since a square root has two values, one positive and the various other negativex2 - 6x + 13 = 0has 2 solutions:x = 3 + √ 4 • iorx = 3 - √ 4 • i

2.3Solvingx2-6x+13 = 0 by the Quadratic Formula.According to the Quadratic Formula,x, the systems forAx2+Bx+C= 0 , wherein A, B and also C space numbers, often called coefficients, is provided by :-B± √B2-4ACx = ————————2A In ours case,A= 1B= -6C= 13 Accordingly,B2-4AC=36 - 52 =-16Applying the quadratic formula : 6 ± √ -16 x=—————2In the set of actual numbers, negative numbers do not have actually square roots. A new set that numbers, referred to as complex, was developed so that an adverse numbers would have actually a square root.

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This numbers room written (a+b*i)Both i and also -i are the square root of minus 1Accordingly,√-16=√16•(-1)=√16•√-1=±√ 16 •i have the right to √ 16 be simplified ?Yes!The prime factorization that 16is2•2•2•2 To have the ability to remove something native under the radical, there need to be 2 instances of that (because we are taking a square i.e. Second root).√ 16 =√2•2•2•2 =2•2•√ 1 =±4 •√ 1 =±4 So currently we room looking at:x=(6±4i )/2Two imaginary services :

x =(6+√-16)/2=3+2i= 3.0000+2.0000ior: x =(6-√-16)/2=3-2i= 3.0000-2.0000i

## Two solutions were discovered :

x =(6-√-16)/2=3-2i= 3.0000-2.0000ix =(6+√-16)/2=3+2i= 3.0000+2.0000i