Tập nghiệm của bất phương trình 2x2-14x+20 0

§2. BẤT PHƯƠNG TRÌNH VÀ HỆ BẤT PHƯƠNG TRÌNH MỘT Ẩn A. KIẾN THỨC CĂN BẢN Bất phương trình một ẩn Bất phương trình ẩn X là mệnh đề chứa biến có dạng f(x)0, Vx B. PHƯƠNG PHÁP GIẢI BÀI TẬP Tìm các giá trị X thoả mãn điều kiện của mỗi bất phương trình sau . 1 , 1 a) - < 1 - —— ; X X +1 c) 2|x|-1+^x-1 < 2* 2x 2 . ? ... „ X -4 X -4x + 3 d) 2Ự1 — X > 3x + —— . x + 4 tfuii a) Điều kiện: ] X e R \ {0; -II X * -1 b) Điều kiện: xz - 4 * 0 X2 - 4x + 3 * 0 X * ±2 X * 1 o X e K \ (1; 3; 2; -2} X * 3 Điều kiện: X * -1 X e \ (-1Ị Điều kiện: p X-0JX_1xe (-ao; 1]\{—4| IX + 4 * 0 X * -4 b) ựl + 2(x-3)2 +V5-4X + X2 <1 Chứng minh các bất phương trinh sau vô nghiệm a) X2 + ựx + 8 < -3 ; Vl + X2 - \Ỉ7 + X2 > 1. éjiải Vì X2 > 0 và yJx + 8 > 0, Vx > -8 nên X2 + Vx + 8 > 0, Vx > -8 Vậy bất phương trình đã cho vô nghiệm. Vì ựl + 2(x - 3)2 > 1 và Võ - 4x + X2 = ^1 + (x - 2)2 > 1 với mọi X nên yỊl + 2(x - 3)2 + Võ - 4x + X2 > 2, Vx e K Vậy bất phương trình đã cho vô nghiệm. Vì Vl + X2 < Vĩ + X2 nên Vl + X2 - Vĩ + X2 < 0. Vx e R Bâ't phương trình đã cho vô nghiệm. Giải thích vì sao các cặp bất phương trình sau tương đương? a) -4x + 1 > 0 và 4x - 1 < 0; b) 2x2 + 5 < 2x - 1 và 2x2 - 2x + 6 < 0; c)x+1>0vàx+1+ 1 > —; d) Vx-1 > X và (2x + 1)v/x — 1 > x(2x +1). x2 + 1 X +1 ốỹúii a) Nhân hai vế bất phương trình thứ nhát với -1 và đổi chiều ta được bâ't phương trình thứ hai (tương đương). b) Chuyển vế và đổi dấu các hạng tử ta được bất phương trình tương đương. 1 c) Cộng vào hai vế bất phương trình với biểu thức X2 +1 không làm thay đổi điều kiện của bất phương trình ta được bất phương trình tương đương, d) Hai bâ’t phương trình có điều kiện chung là X > 1. Trên tập các giá trị này của X thì biểu thức 2x + 1 > 0 nên nhân hai vế bất phương trình thứ nhất với biểu thức 2x + 1 ta được bất phương trình thứ hai (tương đương). 4. Giải các bất phương trinh sau 3x + 1 _ x-2 1-2x . 2 3 < 4 ' a) b) (2x - 1 )(x + 3) - 3x + 1 < (X - 1)(x + 3) + X2 - 5. a) 3x +1 X-2 1-2X 3(3x +1) - 2(x - 2) 1 - 2x n —- — -2 ’ ù L _ 2— < 0 2 3 4 6 4 7x + 7 2x - 1 o < 0 <2 14x + 14 + 6x - 3 < 0 o 20x < -11 11 c _ f 11 X < - ^. Vậy s = -o°;-êê 20 V 20 b) (2x - l)(x + 3) - 3x + 1< (x - l)(x + 3) + X2 - 5 2x2 + 5x-3-3x + 1 1 < -5 vô nghiệm, s = 0. 5. Giải các hệ bất phương trình: a) a) 8x + 3 < 2x + 5 6x + < 4x + 7 7 8x + 3 < 2x + 5; tyZd’z 2x < 7 -1 7 8x + 3 < 4x + 10 22 b) „ 44 2x < — 7 4x < 7 2 3 3X-14 15x-2>2x + 2(x -4) < X < ỳ . Vậy s = -oo; b) 15x - 2 > 2x + 3 13x> 2(x - 4) 3x -14 4x - 16 < 3x -14 X > — 39 X < 2 J- < X < 2. Vậy: s = |-J-;2 I. 39 139 1 c. BÀI TẬP LÀM THÊM Giải các bất phương trình sau: a) 2(x - 1) + X > ^7-3 + 2; . x+2x-2x-1_x c) —— + —— >3 + 7 2 3 4 2 Giải và biện luận các phương trình: a) m2x - 1 > X + m; b) (X + Tã )2 > (x - 72 )2 + 2; (m-1)x 1-x x-1 b) — > -—- - ——- 2(m + 2)> 2 m + 2 3x-1 3(x-2) 5-3X 4x-1 x-1 4-5x 18 >_Ĩ2 9— X + 4m 2x-1 Tìm số nguyên lớn nhất thỏa mãn hệ phương trình: ĩvp iể: X = 4. Xác định m để hệ bất phương trình sau có nghiệm: T)áf} iế: m > - 2.

(2x2 - 14x) + 20 = 0

 3.1     Pull out like factors :

   2x2 - 14x + 20  =   2 • (x2 - 7x + 10) 

Trying to factor by splitting the middle term

 3.2     Factoring  x2 - 7x + 10 

The first term is,  x2  its coefficient is  1 .

The middle term is,  -7x  its coefficient is  -7 .
The last term, "the constant", is  +10 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 10 = 10 

Step-2 : Find two factors of  10  whose sum equals the coefficient of the middle term, which is   -7 .

	-10	+	-1	=	-11
	-5	+	-2	=	-7	That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -5  and  -2 
                     x2 - 5x - 2x - 10Step-4 : Add up the first 2 terms, pulling out like factors :

                    x • (x-5)

              Add up the last 2 terms, pulling out common factors :

                    2 • (x-5)

Step-5 : Add up the four terms of step 4 :

                    (x-2)  •  (x-5)

             Which is the desired factorization

Equation at the end of step  3  :

2 • (x - 2) • (x - 5) = 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 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.

Equations which are never true :

 4.2      Solve :    2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 4.3      Solve  :    x-2 = 0Add  2  to both sides of the equation : 
                      x = 2

Solving a Single Variable Equation :

 4.4      Solve  :    x-5 = 0Add  5  to both sides of the equation : 
                      x = 5

Supplement : Solving Quadratic Equation Directly

Solving  x2-7x+10  = 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   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, 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   3.5000 Plugging into the parabola formula   3.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * 3.50 * 3.50 - 7.0 * 3.50 + 10.0
or   y = -2.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2-7x+10
Axis of Symmetry (dashed)  {x}={ 3.50} 
Vertex at  {x,y} = { 3.50,-2.25} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 2.00, 0.00} 
Root 2 at  {x,y} = { 5.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 5.2     Solving   x2-7x+10 = 0 by Completing The SquareSubtract  10  from both side of the equation :
   x2-7x = -10

Now the clever bit: Take the coefficient of  x , which is  7 , divide by two, giving  7/2 , and finally square it giving  49/4 

Add  49/4  to both sides of the equation :

  On the right hand side we have :

   -10  +  49/4    or,  (-10/1)+(49/4) 

  The common denominator of the two fractions is  4   Adding  (-40/4)+(49/4)  gives  9/4   So adding to both sides we finally get :

   x2-7x+(49/4) = 9/4

Adding  49/4  has completed the left hand side into a perfect square :

   x2-7x+(49/4)  =
   (x-(7/2)) • (x-(7/2))  =
  (x-(7/2))2 Things which are equal to the same thing are also equal to one another. Since

   x2-7x+(49/4) = 9/4 and

   x2-7x+(49/4) = (x-(7/2))2 then, according to the law of transitivity,

   (x-(7/2))2 = 9/4

We'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

   (x-(7/2))2   is

   (x-(7/2))2/2 =
  (x-(7/2))1 =
   x-(7/2)

Now, applying the Square Root Principle to  Eq. #5.2.1  we get:

   x-(7/2) = √ 9/4

Add  7/2  to both sides to obtain:

   x = 7/2 + √ 9/4 Since a square root has two values, one positive and the other negative

   x2 - 7x + 10 = 0

   has two solutions:

  x = 7/2 + √ 9/4

   or

  x = 7/2 - √ 9/4

Note that  √ 9/4 can be written as

  √ 9  / √ 4   which is 3 / 2

Solve Quadratic Equation using the Quadratic Formula

 5.3     Solving    x2-7x+10 = 0 by the Quadratic FormulaAccording to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                     
            - B  ±  √ B2-4AC  x =   ————————                      2A

  In our case,  Accordingly,  Applying the quadratic formula :

               Can  √ 9 be simplified ?

Yes!   The prime factorization of  9   is

   3•3 
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).

√ 9   =  √ 3•3   =
                ±  3 • √ 1   =
                ±  3

So now we are looking at:

           x  =  ( 7 ± 3) / 2

Two real solutions:

x =(7+√9)/2=(7+3)/2= 5.000

or:

x =(7-√9)/2=(7-3)/2= 2.000

Two solutions were found :

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(2x2 - 14x) + 20 = 0

 3.1     Pull out like factors :

   2x2 - 14x + 20  =   2 • (x2 - 7x + 10) 

Trying to factor by splitting the middle term

 3.2     Factoring  x2 - 7x + 10 

The first term is,  x2  its coefficient is  1 .

The middle term is,  -7x  its coefficient is  -7 .
The last term, "the constant", is  +10 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 10 = 10 

Step-2 : Find two factors of  10  whose sum equals the coefficient of the middle term, which is   -7 .

	-10	+	-1	=	-11
	-5	+	-2	=	-7	That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -5  and  -2 
                     x2 - 5x - 2x - 10Step-4 : Add up the first 2 terms, pulling out like factors :

                    x • (x-5)

              Add up the last 2 terms, pulling out common factors :

                    2 • (x-5)

Step-5 : Add up the four terms of step 4 :

                    (x-2)  •  (x-5)

             Which is the desired factorization

Equation at the end of step  3  :

2 • (x - 2) • (x - 5) = 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 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.

Equations which are never true :

 4.2      Solve :    2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Solving a Single Variable Equation :

 4.3      Solve  :    x-2 = 0Add  2  to both sides of the equation : 
                      x = 2

Solving a Single Variable Equation :

 4.4      Solve  :    x-5 = 0Add  5  to both sides of the equation : 
                      x = 5

Supplement : Solving Quadratic Equation Directly

Solving  x2-7x+10  = 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   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, 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   3.5000 Plugging into the parabola formula   3.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * 3.50 * 3.50 - 7.0 * 3.50 + 10.0
or   y = -2.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2-7x+10
Axis of Symmetry (dashed)  {x}={ 3.50} 
Vertex at  {x,y} = { 3.50,-2.25} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = { 2.00, 0.00} 
Root 2 at  {x,y} = { 5.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 5.2     Solving   x2-7x+10 = 0 by Completing The SquareSubtract  10  from both side of the equation :
   x2-7x = -10

Now the clever bit: Take the coefficient of  x , which is  7 , divide by two, giving  7/2 , and finally square it giving  49/4 

Add  49/4  to both sides of the equation :

  On the right hand side we have :

   -10  +  49/4    or,  (-10/1)+(49/4) 

  The common denominator of the two fractions is  4   Adding  (-40/4)+(49/4)  gives  9/4   So adding to both sides we finally get :

   x2-7x+(49/4) = 9/4

Adding  49/4  has completed the left hand side into a perfect square :

   x2-7x+(49/4)  =
   (x-(7/2)) • (x-(7/2))  =
  (x-(7/2))2 Things which are equal to the same thing are also equal to one another. Since

   x2-7x+(49/4) = 9/4 and

   x2-7x+(49/4) = (x-(7/2))2 then, according to the law of transitivity,

   (x-(7/2))2 = 9/4

We'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

   (x-(7/2))2   is

   (x-(7/2))2/2 =
  (x-(7/2))1 =
   x-(7/2)

Now, applying the Square Root Principle to  Eq. #5.2.1  we get:

   x-(7/2) = √ 9/4

Add  7/2  to both sides to obtain:

   x = 7/2 + √ 9/4 Since a square root has two values, one positive and the other negative

   x2 - 7x + 10 = 0

   has two solutions:

  x = 7/2 + √ 9/4

   or

  x = 7/2 - √ 9/4

Note that  √ 9/4 can be written as

  √ 9  / √ 4   which is 3 / 2

Solve Quadratic Equation using the Quadratic Formula

 5.3     Solving    x2-7x+10 = 0 by the Quadratic FormulaAccording to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                     
            - B  ±  √ B2-4AC  x =   ————————                      2A

  In our case,  Accordingly,  Applying the quadratic formula :

               Can  √ 9 be simplified ?

Yes!   The prime factorization of  9   is

   3•3 
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).

√ 9   =  √ 3•3   =
                ±  3 • √ 1   =
                ±  3

So now we are looking at:

           x  =  ( 7 ± 3) / 2

Two real solutions:

x =(7+√9)/2=(7+3)/2= 5.000

or:

x =(7-√9)/2=(7-3)/2= 2.000

Two solutions were found :