Math, asked by adib646, 7 months ago

what is the completing square of (7-3x)(x+2)=4?​

Answers

Answered by paroshnee18
1

Answer:

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Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

(7 - 3x) • (x + 2) -  4  = 0

Factoring  -3x2+x+10  

The first term is,  -3x2  its coefficient is  -3 .

The middle term is,  +x  its coefficient is  1 .

The last term, "the constant", is  +10  

Step-1 : Multiply the coefficient of the first term by the constant   -3 • 10 = -30  

Step-2 : Find two factors of  -30  whose sum equals the coefficient of the middle term, which is   1 .

     -30    +    1    =    -29  

     -15    +    2    =    -13  

     -10    +    3    =    -7  

     -6    +    5    =    -1  

     -5    +    6    =    1    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -5  and  6  

                    -3x2 - 5x + 6x + 10

Step-4 : Add up the first 2 terms, pulling out like factors :

                   -x • (3x+5)

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

                   2 • (3x+5)

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

                   (-x+2)  •  (3x+5)

            Which is the desired factorization

(3x + 5) • (2 - x)  = 0

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.

Solve  :    3x+5 = 0  

Subtract  5  from both sides of the equation :  

                     3x = -5

Divide both sides of the equation by 3:

                    x = -5/3 = -1.667

Solve  :    -x+2 = 0  

Subtract  2  from both sides of the equation :  

                     -x = -2

Multiply both sides of the equation by (-1) :  x = 2

Solving    -3x2+x+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

Find the Vertex of   y = -3x2+x+10

Parabolas have a highest or a lowest point called the Vertex .  

Solve Quadratic Equation by Completing The Square

4.2     Solving   -3x2+x+10 = 0 by Completing The Square .

Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:

3x2-x-10 = 0  Divide both sides of the equation by  3  to have 1 as the coefficient of the first term :

  x2-(1/3)x-(10/3) = 0

Add  10/3  to both side of the equation :

  x2-(1/3)x = 10/3

Now the clever bit: Take the coefficient of  x , which is  1/3 , divide by two, giving  1/6 , and finally square it giving  1/36  

Add  1/36  to both sides of the equation :

 On the right hand side we have :

  10/3  +  1/36   The common denominator of the two fractions is  36   Adding  (120/36)+(1/36)  gives  121/36  

 So adding to both sides we finally get :

  x2-(1/3)x+(1/36) = 121/36

Adding  1/36  has completed the left hand side into a perfect square :

  x2-(1/3)x+(1/36)  =

  (x-(1/6)) • (x-(1/6))  =

 (x-(1/6))2

Things which are equal to the same thing are also equal to one another. Since

  x2-(1/3)x+(1/36) = 121/36 and

  x2-(1/3)x+(1/36) = (x-(1/6))2

then, according to the law of transitivity,

  (x-(1/6))2 = 121/36

We'll refer to this Equation as  Eq. #4.2.1  

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

Note that the square root of

  (x-(1/6))2   is

  (x-(1/6))2/2 =

 (x-(1/6))1 =

  x-(1/6)

Now, applying the Square Root Principle to  Eq. #4.2.1  we get:

  x-(1/6) = √ 121/36

Add  1/6  to both sides to obtain:

  x = 1/6 + √ 121/36

Since a square root has two values, one positive and the other negative

  x2 - (1/3)x - (10/3) = 0

  has two solutions:

 x = 1/6 + √ 121/36

  or

 x = 1/6 - √ 121/36

Note that  √ 121/36 can be written as

 √ 121  / √ 36   which is 11 / 6

Solve Quadratic Equation using the Quadratic Formula

4.3     Solving    -3x2+x+10 = 0 by the Quadratic Formula .

According 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,  A   =     -3

                     B   =    1

                     C   =   10

Accordingly,  B2  -  4AC   =

                    1 - (-120) =

                    121

Apply the quadratic formula :

              -1 ± √ 121

  x  =    ——————

                     -6

Can  √ 121 be simplified ?

Yes!   The prime factorization of  121   is

  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:

          x  =  ( -1 ± 11) / -6

Two real solutions:

x =(-1+√121)/-6=(1-11)/6= -1.667  / x =(-1-√121)/-6=(1+11)/6= 2.000

Two solutions were found :

x = 2

x = -5/3 = -1.667

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