### 1/4x^2-1/2x=1

This solution deals with adding, subtracting and finding the least common multiple.

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## Step by Step Solution

### Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 1/4*x^2-1/2*x-(1)=0## Step by step solution :

## Step 1 :

1 Simplify — 2Equation at the end of step 1 : 1 1 ((—•(x2))-(—•x))-1 = 0 4 2

## Step 2 :

1 Simplify — 4Equation at the end of step 2 : 1 x ((— • x2) - —) - 1 = 0 4 2## Step 3 :

Equation at the end of step 3 : x2 x (—— - —) - 1 = 0 4 2## Step 4 :

Calculating the Least Common Multiple :4.1 Find the Least Common Multiple The left denominator is : 4 The right denominator is : 2Number of times each prime factorappears in the factorization of:PrimeFactorLeftDenominatorRightDenominatorL.C.M = Max{Left,Right}2 | 2 | 1 | 2 |

Product of allPrime Factors | 4 | 2 | 4 |

Least Common Multiple: 4

Calculating Multipliers :4.2 Calculate multipliers for the two fractions Denote the Least Common Multiple by L.C.M Denote the Left Multiplier by Left_M Denote the Right Multiplier by Right_M Denote the Left Deniminator by L_Deno Denote the Right Multiplier by R_DenoLeft_M=L.C.M/L_Deno=1Right_M=L.C.M/R_Deno=2

Making Equivalent Fractions :4.3 Rewrite the two fractions into equivalent fractionsTwo fractions are called equivalent if they have the same numeric value. For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well. To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

L. Mult. • L. Num. x2 —————————————————— = —— L.C.M 4 R. Mult. • R. Num. x • 2 —————————————————— = ————— L.C.M 4 Adding fractions that have a common denominator :4.4 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:

x2 - (x • 2) x2 - 2x ———————————— = ——————— 4 4 Equation at the end of step 4 : (x2 - 2x) ————————— - 1 = 0 4

## Step 5 :

Rewriting the whole as an Equivalent Fraction :5.1Subtracting a whole from a fraction Rewrite the whole as a fraction using 4 as the denominator :1 1 • 4 1 = — = ————— 1 4 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

## Step 6 :

Pulling out like terms :6.1 Pull out like factors:x2 - 2x=x•(x - 2)Adding fractions that have a common denominator :6.2 Adding up the two equivalent fractions

x • (x-2) - (4) x2 - 2x - 4 ——————————————— = ——————————— 4 4 Trying to factor by splitting the middle term6.3Factoring x2 - 2x - 4 The first term is, x2 its coefficient is 1.The middle term is, -2x its coefficient is -2.The last term, "the constant", is -4Step-1 : Multiply the coefficient of the first term by the constant 1•-4=-4Step-2 : Find two factors of -4 whose sum equals the coefficient of the middle term, which is -2.

-4 | + | 1 | = | -3 | ||

-2 | + | 2 | = | 0 | ||

-1 | + | 4 | = | 3 |

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

Equation at the end of step 6 : x2 - 2x - 4 ——————————— = 0 4

## Step 7 :

When a fraction equals zero :7.1 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:x2-2x-4 ——————— • 4 = 0 • 4 4 Now, on the left hand side, the 4 cancels out the denominator, while, on the right hand side, zero times anything is still zero.The equation now takes the shape:x2-2x-4=0

Parabola, Finding the Vertex:7.2Find the Vertex ofy = x2-2x-4Parabolas 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 1.0000Plugging into the parabola formula 1.0000 for x we can calculate the y-coordinate:y = 1.0 * 1.00 * 1.00 - 2.0 * 1.00 - 4.0 or y = -5.000

Parabola, Graphing Vertex and X-Intercepts :Root plot for : y = x2-2x-4 Axis of Symmetry (dashed) {x}={ 1.00} Vertex at {x,y} = { 1.00,-5.00} x-Intercepts (Roots) : Root 1 at {x,y} = {-1.24, 0.00} Root 2 at {x,y} = { 3.24, 0.00}

Solve Quadratic Equation by Completing The Square7.3Solvingx2-2x-4 = 0 by Completing The Square.Add 4 to both side of the equation : x2-2x = 4Now the clever bit: Take the coefficient of x, which is 2, divide by two, giving 1, and finally square it giving 1Add 1 to both sides of the equation :On the right hand side we have:4+1or, (4/1)+(1/1)The common denominator of the two fractions is 1Adding (4/1)+(1/1) gives 5/1So adding to both sides we finally get:x2-2x+1 = 5Adding 1 has completed the left hand side into a perfect square :x2-2x+1=(x-1)•(x-1)=(x-1)2 Things which are equal to the same thing are also equal to one another. Sincex2-2x+1 = 5 andx2-2x+1 = (x-1)2 then, according to the law of transitivity,(x-1)2 = 5We"ll refer to this Equation as Eq.

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#7.3.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of(x-1)2 is(x-1)2/2=(x-1)1=x-1Now, applying the Square Root Principle to Eq.#7.3.1 we get:x-1= √ 5 Add 1 to both sides to obtain:x = 1 + √ 5 Since a square root has two values, one positive and the other negativex2 - 2x - 4 = 0has two solutions:x = 1 + √ 5 orx = 1 - √ 5