Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
-(24)/(y^(2)+3*y-28)-(-(2*y)/(y^(2)+7))=0
Step by step solution :
Step 1 :
2y Simplify —————— y2 + 7
Polynomial Roots Calculator :
1.1 Find roots (zeroes) of : F(y) = y2 + 7
Polynomial Roots Calculator is a set of methods aimed at finding values ofyfor which F(y)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers y which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational numberP/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient
In this case, the Leading Coefficient is 1 and the Trailing Constant is 7. The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,7 Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | 8.00 | ||||||
-7 | 1 | -7.00 | 56.00 | ||||||
1 | 1 | 1.00 | 8.00 | ||||||
7 | 1 | 7.00 | 56.00 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 1 :
24 2y (0-——————————————)-(0-————) = 0 (((y2)+3y)-28) y2+7
Step 2 :
24 Simplify ———————————— y2 + 3y - 28
Trying to factor by splitting the middle term
2.1Factoring y2 + 3y - 28
The first term is, y2 its coefficient is 1.
The middle term is, +3y its coefficient is 3.
The last term, "the constant", is -28
Step-1 : Multiply the coefficient of the first term by the constant 1•-28=-28
Step-2 : Find two factors of -28 whose sum equals the coefficient of the middle term, which is 3.
-28 | + | 1 | = | -27 | ||
-14 | + | 2 | = | -12 | ||
-7 | + | 4 | = | -3 | ||
-4 | + | 7 | = | 3 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step2above, -4 and 7
y2 - 4y+7y - 28
Step-4 : Add up the first 2 terms, pulling out like factors:
y•(y-4)
Add up the last 2 terms, pulling out common factors:
7•(y-4)
Step-5:Add up the four terms of step4:
(y+7)•(y-4)
Which is the desired factorization
Equation at the end of step 2 :
24 -2y (0 - —————————————————) - —————— = 0 (y + 7) • (y - 4) y2 + 7
Step 3 :
Calculating the Least Common Multiple :
3.1 Find the Least Common Multiple
The left denominator is : (y+7)•(y-4)
The right denominator is : y2+7
Algebraic Factor | Left Denominator | Right Denominator | L.C.M = Max {Left,Right} |
---|---|---|---|
y+7 | 1 | 0 | 1 |
y-4 | 1 | 0 | 1 |
y2+7 | 0 | 1 | 1 |
Least Common Multiple:
(y+7)•(y-4)•(y2+7)
Calculating Multipliers :
3.2 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_Deno
Left_M=L.C.M/L_Deno=y2+7
Right_M=L.C.M/R_Deno=(y+7)•(y-4)
Making Equivalent Fractions :
3.3 Rewrite the two fractions into equivalent fractions
Two 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. -24 • (y2+7) —————————————————— = —————————————————————— L.C.M (y+7) • (y-4) • (y2+7) R. Mult. • R. Num. -2y • (y+7) • (y-4) —————————————————— = —————————————————————— L.C.M (y+7) • (y-4) • (y2+7)
Adding fractions that have a common denominator :
3.4 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
-24 • (y2+7) - (-2y • (y+7) • (y-4)) 2y3 - 18y2 - 56y - 168 ———————————————————————————————————— = ———————————————————————————— (y+7) • (y-4) • (y2+7) (y + 7) • (y - 4) • (y2 + 7)
Step 4 :
Pulling out like terms :
Checking for a perfect cube :
4.2y3 - 9y2 - 28y - 84 is not a perfect cube
Trying to factor by pulling out :
4.3 Factoring: y3 - 9y2 - 28y - 84
Thoughtfully split the expression at hand into groups, each group having two terms:
Group 1: y3 - 9y2
Group 2: -28y - 84
Pull out from each group separately :
Group 1: (y - 9) • (y2)
Group 2: (y + 3) • (-28)
Bad news !! Factoring by pulling out fails : The groups have no common factor and can not be added up to form a multiplication.
Polynomial Roots Calculator :
4.4 Find roots (zeroes) of : F(y) = y3 - 9y2 - 28y - 84
See theory in step 1.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is -84. The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,7 ,12 ,14 ,21 ,28 , etc Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | -66.00 | ||||||
-2 | 1 | -2.00 | -72.00 | ||||||
-3 | 1 | -3.00 | -108.00 | ||||||
-4 | 1 | -4.00 | -180.00 | ||||||
-6 | 1 | -6.00 | -456.00 |
Note - For tidiness, printing of 15 checks which found no root was suppressed
Polynomial Roots Calculator found no rational roots
Equation at the end of step 4 :
2 • (y3 - 9y2 - 28y - 84) ———————————————————————————— = 0 (y + 7) • (y - 4) • (y2 + 7)
Step 5 :
When a fraction equals zero :
5.1 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:
2•(y3-9y2-28y-84) —————————————————— • (y+7)•(y-4)•(y2+7) = 0 • (y+7)•(y-4)•(y2+7) (y+7)•(y-4)•(y2+7)
Now, on the left hand side, the (y+7)•(y-4)•(y2+7) cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape:
2 • (y3-9y2-28y-84)=0
Equations which are never true:
5.2Solve:2=0
This equation has no solution.
A a non-zero constant never equals zero.
Cubic Equations:
5.3Solvey3-9y2-28y-84 = 0
Future releases of Tiger-Algebra will solve equations of the third degree directly.
Meanwhile we will use the Bisection method to approximate one real solution.
Approximating a root using the Bisection Method :
We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).
The function is F(y) = y3 - 9y2 - 28y - 84
Aty= 11.00 F(y) is equal to -150.00
Aty= 12.00 F(y) is equal to 12.00
Intuitively we feel, and justly so, that since F(y) is negative on one side of the interval, and positive on the other side then, somewhere inside this interval, F(y) is zero
Procedure :
(1) Find a point "Left" where F(Left) < 0
(2) Find a point 'Right' where F(Right) > 0
(3) Compute 'Middle' the middle point of the interval [Left,Right]
(4) Calculate Value = F(Middle)
(5) If Value is close enough to zero goto Step (7)
Else :
If Value < 0 then : Left <- Middle
If Value > 0 then : Right <- Middle
(6) Loop back to Step (3)
(7) Done!! The approximation found is Middle
Follow Middle movements to understand how it works :
Left Value(Left) Right Value(Right)11.000000000 -150.000000000 12.000000000 12.000000000 0.000000000 -84.000000000 12.000000000 12.000000000 6.000000000 -360.000000000 12.000000000 12.000000000 9.000000000 -336.000000000 12.000000000 12.00000000010.500000000 -212.625000000 12.000000000 12.00000000011.250000000 -114.234375000 12.000000000 12.00000000011.625000000 -54.755859375 12.000000000 12.00000000011.812500000 -22.307373047 12.000000000 12.00000000011.906250000 -5.388519287 12.000000000 12.00000000011.906250000 -5.388519287 11.953125000 3.24672317511.929687500 -1.085613728 11.953125000 3.24672317511.929687500 -1.085613728 11.941406250 1.07687097811.935546875 -0.005291708 11.941406250 1.07687097811.935546875 -0.005291708 11.938476562 0.53555947611.935546875 -0.005291708 11.937011719 0.26507635411.935546875 -0.005291708 11.936279297 0.12987794211.935546875 -0.005291708 11.935913086 0.06228952211.935546875 -0.005291708 11.935729980 0.02849800811.935546875 -0.005291708 11.935638428 0.01160292511.935546875 -0.005291708 11.935592651 0.00315555311.935569763 -0.001068092 11.935592651 0.00315555311.935569763 -0.001068092 11.935581207 0.00104372711.935575485 -0.000012183 11.935581207 0.00104372711.935575485 -0.000012183 11.935578346 0.00051577211.935575485 -0.000012183 11.935576916 0.00025179411.935575485 -0.000012183 11.935576200 0.00011980611.935575485 -0.000012183 11.935575843 0.00005381111.935575485 -0.000012183 11.935575664 0.00002081411.935575485 -0.000012183 11.935575575 0.000004315
Next Middle will get us close enough to zero:
F(11.935575552) is 0.000000191
The desired approximation of the solution is:
y ≓ 11.935575552
Note, ≓ is the approximation symbol
Supplement : Solving Quadratic Equation Directly
Solving y2 + 3y - 28 = 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:
6.1Find the Vertex oft = y2+3y-28Parabolas 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 "t" 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,Ay2+By+C,the y-coordinate of the vertex is given by -B/(2A). In our case the y coordinate is -1.5000Plugging into the parabola formula -1.5000 for y we can calculate the t-coordinate:
t = 1.0 * -1.50 * -1.50 + 3.0 * -1.50 - 28.0
or t = -30.250
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : t = y2+3y-28
Axis of Symmetry (dashed) {y}={-1.50}
Vertex at {y,t} = {-1.50,-30.25}
y-Intercepts (Roots) :
Root 1 at {y,t} = {-7.00, 0.00}
Root 2 at {y,t} = { 4.00, 0.00}
Solve Quadratic Equation by Completing The Square
6.2Solvingy2+3y-28 = 0 by Completing The Square.Add 28 to both side of the equation :
y2+3y = 28
Now the clever bit: Take the coefficient of y, which is 3, divide by two, giving 3/2, and finally square it giving 9/4
Add 9/4 to both sides of the equation :
On the right hand side we have:
28+9/4or, (28/1)+(9/4)
The common denominator of the two fractions is 4Adding (112/4)+(9/4) gives 121/4
So adding to both sides we finally get:
y2+3y+(9/4) = 121/4
Adding 9/4 has completed the left hand side into a perfect square :
y2+3y+(9/4)=
(y+(3/2))•(y+(3/2))=
(y+(3/2))2
Things which are equal to the same thing are also equal to one another. Since
y2+3y+(9/4) = 121/4 and
y2+3y+(9/4) = (y+(3/2))2
then, according to the law of transitivity,
(y+(3/2))2 = 121/4
We'll refer to this Equation as Eq. #6.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(y+(3/2))2 is
(y+(3/2))2/2=
(y+(3/2))1=
y+(3/2)
Now, applying the Square Root Principle to Eq.#6.2.1 we get:
y+(3/2)= √ 121/4
Subtract 3/2 from both sides to obtain:
y = -3/2 + √ 121/4
Since a square root has two values, one positive and the other negative
y2 + 3y - 28 = 0
has two solutions:
y = -3/2 + √ 121/4
or
y = -3/2 - √ 121/4
Note that √ 121/4 can be written as
√121 / √4which is 11 / 2
Solve Quadratic Equation using the Quadratic Formula
6.3Solvingy2+3y-28 = 0 by the Quadratic Formula.According to the Quadratic Formula,y, the solution forAy2+By+C= 0 , where A, B and C are numbers, often called coefficients, is given by :
-B± √B2-4AC
y = ————————
2A In our case,A= 1
B= 3
C=-28 Accordingly,B2-4AC=
9 - (-112) =
121Applying the quadratic formula :
-3 ± √ 121
y=——————
2Can √ 121 be simplified ?
Yes!The prime factorization of 121is
11•11
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).
√ 121 =√11•11 =
±11 •√ 1 =
±11
So now we are looking at:
y=(-3±11)/2
Two real solutions:
y =(-3+√121)/2=(-3+11)/2= 4.000
or:
y =(-3-√121)/2=(-3-11)/2= -7.000
One solution was found :
y ≓ 11.935575552