Question

3(x-4)^-5(x-4)=12 find the roots of quadratic equation by factorisation easy method

Answer 1

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Let x-4 = y
Then the equation becomes 3y²-5y-12=0
3y²-9y+4y-12 = 0
3y(y-3)+4(y-3) = 0
(y-3)(3y+4) = 0
y = 3 or -4/3
x-4 = 3 or -4/3
x = 7 or 8/3

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Answer 2

(3•((x-4)2))-5•(x-4))-12 = 0

Step  2  :

Equation at the end of step  2  :

(3 • (x - 4)2 - 5 • (x - 4)) - 12 = 0

Step  3  :

Trying to factor by splitting the middle term

 3.1     Factoring  3x2-29x+56 

The first term is,  3x2  its coefficient is  3 .
The middle term is,  -29x  its coefficient is  -29 .
The last term, "the constant", is  +56 

Step-1 : Multiply the coefficient of the first term by the constant   3 • 56 = 168 

Step-2 : Find two factors of  168  whose sum equals the coefficient of the middle term, which is   -29 .

     -168   +   -1   =   -169     -84   +   -2   =   -86     -56   +   -3   =   -59     -42   +   -4   =   -46     -28   +   -6   =   -34     -24   +   -7   =   -31     -21   +   -8   =   -29   That's it


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -21  and  -8 
                     3x2 - 21x - 8x - 56

Step-4 : Add up the first 2 terms, pulling out like factors :
                    3x • (x-7)
              Add up the last 2 terms, pulling out common factors :
                    8 • (x-7)
Step-5 : Add up the four terms of step 4 :
                    (3x-8)  •  (x-7)
             Which is the desired factorization

Equation at the end of step  3  :

(x - 7) • (3x - 8) = 0

Step  4  :

Theory - Roots of a product :

 4.1    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 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 4.2      Solve  :    x-7 = 0 

 Add  7  to both sides of the equation : 
                      x = 7 

Solving a Single Variable Equation :

 4.3      Solve  :    3x-8 = 0 

 Add  8  to both sides of the equation : 
                      3x = 8 
Divide both sides of the equation by 3:
                     x = 8/3 = 2.667 

Supplement : Solving Quadratic Equation Directly

Solving  3x2-29x+56  = 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 = 3x2-29x+56

Parabolas 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, 3 , 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   4.8333  

 Plugging into the parabola formula   4.8333  for  x  we can calculate the  y -coordinate : 
  y = 3.0 * 4.83 * 4.83 - 29.0 * 4.83 + 56.0 
or   y = -14.083

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 3x2-29x+56
Axis of Symmetry (dashed)  {x}={ 4.83} 
Vertex at  {x,y} = { 4.83,-14.08}  
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 2.67, 0.00} 
Root 2 at  {x,y} = { 7.00, 0.00}