Solve Quadraticequations N(n-1)/2=78 Tiger Algebra Solver

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : n*(n-1)/2-(78)=0

Step by step solution :

Step 1 :

n - 1 Simplify ————— 2

Equation at the end of step 1 :

(n - 1) (n • ———————) - 78 = 0 2

Step 2 :

Equation at the end of step 2 :

n • (n - 1) ——————————— - 78 = 0 2

Step 3 :

Rewriting the whole as an Equivalent Fraction :

3.1 Subtracting a whole from a fraction Rewrite the whole as a fraction using 2 as the denominator :

78 78 • 2 78 = —— = —————— 1 2

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

3.2 Adding up the two equivalent fractions Add the two equivalent fractions which now have a common denominatorCombine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

n • (n-1) - (78 • 2) n2 - n - 156 ———————————————————— = ———————————— 2 2

Trying to factor by splitting the middle term

3.3 Factoring n2 - n - 156 The first term is, n2 its coefficient is 1 .The middle term is, -n its coefficient is -1 .The last term, "the constant", is -156 Step-1 : Multiply the coefficient of the first term by the constant 1 • -156 = -156 Step-2 : Find two factors of -156 whose sum equals the coefficient of the middle term, which is -1 .

-156	+	1	=	-155
-78	+	2	=	-76
-52	+	3	=	-49
-39	+	4	=	-35
-26	+	6	=	-20
-13	+	12	=	-1	That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -13 and 12 n2 - 13n + 12n - 156Step-4 : Add up the first 2 terms, pulling out like factors : n • (n-13) Add up the last 2 terms, pulling out common factors : 12 • (n-13) Step-5 : Add up the four terms of step 4 : (n+12) • (n-13) Which is the desired factorization

Equation at the end of step 3 :

(n + 12) • (n - 13) ——————————————————— = 0 2

Step 4 :

When a fraction equals zero :

4.1 When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.Here's how:

(n+12)•(n-13) ————————————— • 2 = 0 • 2 2

Now, on the left hand side, the 2 cancels out the denominator, while, on the right hand side, zero times anything is still zero.The equation now takes the shape : (n+12) • (n-13) = 0

Theory - Roots of a product :

4.2 A product of several terms equals zero.When a product of two or more terms equals zero, then at least one of the terms must be zero.We shall now solve each term = 0 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

4.3 Solve : n+12 = 0Subtract 12 from both sides of the equation : n = -12

Solving a Single Variable Equation :

4.4 Solve : n-13 = 0Add 13 to both sides of the equation : n = 13

Supplement : Solving Quadratic Equation Directly

Solving n2-n-156 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

5.1 Find the Vertex of y = n2-n-156Parabolas 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,An2+Bn+C,the n -coordinate of the vertex is given by -B/(2A) . In our case the n coordinate is 0.5000 Plugging into the parabola formula 0.5000 for n we can calculate the y -coordinate : y = 1.0 * 0.50 * 0.50 - 1.0 * 0.50 - 156.0 or y = -156.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = n2-n-156 Axis of Symmetry (dashed) {n}={ 0.50} Vertex at {n,y} = { 0.50,-156.25} n -Intercepts (Roots) : Root 1 at {n,y} = {-12.00, 0.00} Root 2 at {n,y} = {13.00, 0.00}

Solve Quadratic Equation by Completing The Square

5.2 Solving n2-n-156 = 0 by Completing The Square . Add 156 to both side of the equation : n2-n = 156Now the clever bit: Take the coefficient of n , which is 1 , divide by two, giving 1/2 , and finally square it giving 1/4 Add 1/4 to both sides of the equation : On the right hand side we have : 156 + 1/4 or, (156/1)+(1/4) The common denominator of the two fractions is 4 Adding (624/4)+(1/4) gives 625/4 So adding to both sides we finally get : n2-n+(1/4) = 625/4Adding 1/4 has completed the left hand side into a perfect square : n2-n+(1/4) = (n-(1/2)) • (n-(1/2)) = (n-(1/2))2 Things which are equal to the same thing are also equal to one another. Since n2-n+(1/4) = 625/4 and n2-n+(1/4) = (n-(1/2))2 then, according to the law of transitivity, (n-(1/2))2 = 625/4We'll refer to this Equation as Eq. #5.2.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of (n-(1/2))2 is (n-(1/2))2/2 = (n-(1/2))1 = n-(1/2)Now, applying the Square Root Principle to Eq. #5.2.1 we get: n-(1/2) = √ 625/4 Add 1/2 to both sides to obtain: n = 1/2 + √ 625/4 Since a square root has two values, one positive and the other negative n2 - n - 156 = 0 has two solutions: n = 1/2 + √ 625/4 or n = 1/2 - √ 625/4 Note that √ 625/4 can be written as √ 625 / √ 4 which is 25 / 2

Solve Quadratic Equation using the Quadratic Formula

5.3 Solving n2-n-156 = 0 by the Quadratic Formula . According to the Quadratic Formula, n , the solution for An2+Bn+C = 0 , where A, B and C are numbers, often called coefficients, is given by : - B ± √ B2-4AC n = ———————— 2A In our case, A = 1 B = -1 C = -156 Accordingly, B2 - 4AC = 1 - (-624) = 625Applying the quadratic formula : 1 ± √ 625 n = ————— 2Can √ 625 be simplified ?Yes! The prime factorization of 625 is 5•5•5•5 To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).√ 625 = √ 5•5•5•5 =5•5•√ 1 = ± 25 • √ 1 = ± 25 So now we are looking at: n = ( 1 ± 25) / 2Two real solutions:n =(1+√625)/2=(1+25)/2= 13.000 or:n =(1-√625)/2=(1-25)/2= -12.000

Two solutions were found :

1. n = 13
2. n = -12