\(\Delta=\left(-5\right)^2-4.1.\left(-4500\right)=25+18000=18025>0\)

\(\Rightarrow\)Phương trình có 2 nghiệm phân biệt là:

\(x_1=\frac{-5-\sqrt{18025}}{2}=\frac{-5-5\sqrt{721}}{2}\)và \(x_2=\frac{-5+\sqrt{18025}}{2}=\frac{-5+5\sqrt{721}}{2}\)

Vậy \(x=\frac{-5\pm5\sqrt{721}}{2}\)

1.

The first term is,  x²  its coefficient is  1 .
The middle term is,  +5x  its coefficient is  5 .
The last term, "the constant", is  -4500 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -4500 = -4500 

Step-2 : Find two factors of  -4500  whose sum equals the coefficient of the middle term, which is   5 .

For tidiness, printing of 30 lines which failed to find two such factors, was suppressed

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

2.2.1      Find the Vertex of   y = x²+5x-4500

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 veral 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  -2.5000  

 Plugging into the parabola formula  -2.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * -2.50 * -2.50 + 5.0 * -2.50 - 4500.0
or   y = -4506.250

3.

Root plot for :  y = x²+5x-4500
Axis of Symmetry (dashed)  {x}={-2.50} 
Vertex at  {x,y} = {-2.50,-4506.25} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-69.63, 0.00} 
Root 2 at  {x,y} = {64.63, 0.00} 

Solve Quadra Equation by Completing The Square

 2.2     Solving   x²+5x-4500 = 0 by Completing The Square .

 Add  4500  to both side of the equation :
   x²+5x = 4500

Now the clever bit: Take the coefficient of  x , which is  5 , divide by two, giving  5/2 , and finally square it giving  25/4 

Add  25/4  to both sides of the equation :
  On the right hand side we have :
   4500  +  25/4    or,  (4500/1)+(25/4) 
  The common denominator of the two fractions is  4   Adding  (18000/4)+(25/4)  gives  18025/4 
  So adding to both sides we finally get :
   x²+5x+(25/4) = 18025/4

Adding  25/4  has completed the left hand side into a perfect square :
   x²+5x+(25/4)  =
   (x+(5/2)) • (x+(5/2))  =
  (x+(5/2))²
Things which are equal to the same thing are also equal to one another. Since
   x²+5x+(25/4) = 18025/4 and
   x²+5x+(25/4) = (x+(5/2))²
then, according to the law of transitivity,
   (x+(5/2))² = 18025/4

We'll refer to this Equation as  Eq. #2.2.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x+(5/2))²   is
   (x+(5/2))^2/2 =
  (x+(5/2))¹ =
   x+(5/2)

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
   x+(5/2) = √ 18025/4

Subtract  5/2  from both sides to obtain:
   x = -5/2 + √ 18025/4

Since a square root has two values, one positive and the other negative
   x² + 5x - 4500 = 0
   has two solutions:
  x = -5/2 + √ 18025/4
   or
  x = -5/2 - √ 18025/4

Note that  √ 18025/4 can be written as
  √ 18025  / √ 4   which is √ 18025  / 2

4. 2.3     Solving    x²+5x-4500 = 0 by the Quadra Formula .

 According to the Quadra 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,  A   =     1
                      B   =    5
                      C   =  -4500

Accordingly,  B²  -  4AC   =
                     25 - (-18000) =
                     18025

Applying the quadra formula :

               -5 ± √ 18025
   x  =    ———————
                        2

Can  √ 18025 be simplified ?

Yes!   The prime factorization of  18025   is
   5•5•7•103 
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).

√ 18025   =  √ 5•5•7•103   =
                ±  5 • √ 721

  √ 721   , rounded to 4 decimal digits, is  26.8514
 So now we are looking at:
           x  =  ( -5 ± 5 •  26.851 ) / 2

Two real solutions:

 x =(-5+√18025)/2=(-5+5√ 721 )/2= 64.629

or:

 x =(-5-√18025)/2=(-5-5√ 721 )/2= -69.629

a) \(\Rightarrow x\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

a, 5\(x\)(\(x^2\) - 9) = 0

    \(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\) 

Vậy \(x\) \(\in\) { -3; 0; 3}

b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0

    3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0

    (\(x+3\))( 3 - \(x\)) = 0

     \(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)

Vậy \(x\) \(\in\){ -3; 3}

c, \(x^2\) - 9\(x\) - 10 = 0

   \(x^2\) + \(x\) - 10\(x\)  - 10 = 0

   \(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0

        (\(x+1\))(\(x-10\)) = 0

         \(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)

           \(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

Vậy \(x\) \(\in\){ -1; 10}

a) 5x(x2-9)=0

=> TH1 5x=0

      <=> x= 0

     TH2: 2x-9=0

       <=> 2x=9

       <=> x = \(\dfrac{9}{2}\)

b, 3(x+3) - x²- 3x = 0   

<=> 3x + 9 - x² -3x = 0

<=> - x² +9 = 0

<=> - x² = -9

 <=> x = 3

c, x² -9x -10 = 0

<=> x² -x + 10x -10 = 0

<=> x(x-1)+10(x-1)=0     

 <=> (x-1)(x+10)=0

=> TH1: x-1=0

       <=> x=1

      TH2: x +10=0

      <=>  x=-10

Bạn xem đã viết đúng đề chưa vậy?

`a)5x(x-1)-(x+2)(5x-7)=6`

`<=>5x^2-5x-(5x^2-7x+10x-14)=6`

`<=>5x^2-5x-(5x^2+3x-14)=6`

`<=>-8x+14=6`

`<=>8x=8<=>x=1`

Vậy `x=1`

`b)(x+2)^2-(x^2-4)=0`

`<=>x^2+4x+4-x^2+4=0`

`<=>4x+8=0`

`<=>4x=-8`

`<=>x=-2`

Vậy `x=-2`

a)x=5/2

b)x=-2

a)5x.(x-1)-(x+2).(5x-7)=6

<=> 5x²-5x-(5x²-7x+10x-14)=6

<=>  5x²-5x-5x²+7x-10x+14=6

<=> -8x+14=6

<=> -8x=-8 => x=1

Vậy x=1

b) (x+2)²-(x²-4)=0

<=> (x+2)²-(x²-2²)=0 <=> (x+2)²-(x-2)(x+2)=0

<=> (x+2)[(x+2)-(x-2)]=0

<=> (x+2)(x+2-x+2)=0

<=> (x+2).4=0

=> x+2=0

=> x=-2

Vậy x=-2

a: \(x^2-4x=3\left(x-4\right)\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

b: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

a) pt <=> (x - 4)(x - 3) = 0

<=> x = 4 hoặc x = 3

b) pt <=> (x - 8)(x + 3) = 0

<=> x = 8 hoặc x = -3

Pt <=> (2-x)(x+5)=0

<=> x=2 hoặc x=-5

\(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)