Question

3(x – 9)2 + 15 = 195 Solving quadratics with square root

Answer 1

Answer:

$x - \frac{6 - \sqrt{80} }{2} = 3 - \sqrt[2]{5} = 1.472 \\ x - \frac{6 + \sqrt{80} }{2} = 3 - \sqrt[2]{5} = 7.472$

Step-by-step explanation:

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

${(3x - 9)}^{2} \ + 15 - (195) = 0$  

Step 1 :Evaluate :  (3x-9)2   =  9x^2-54x+81   =   3 • (3x^2 - 18x - 43)    

Step 2: Pull out like factors :

${9x}^{2} - 54x - 99 = 9•( {x}^{2} - 6x - 11)$

Step 3: Trying to factor by splitting the middle term

   Factoring   ${x}^{2} - 6x - 11$

The first term is,  ${x}^{2}$  its coefficient is  1 .

The middle term is,  ${ - 6x}$ its coefficient is -6 .

The last term, "the constant", is $- 11$

 Step 4: Multiply the coefficient of the first term by the constant  

$1 \: • -11 = -11$

Step-5 : Find two factors of -11  whose sum equals the coefficient of the middle term, which is -6. 

$- 11 + 1 = - 10 \\ - 1 + 11 = 10$

$9 • ( {x}^{2} - 6x - 11) = 0$

Step 6: 

  Solve :   

$9 = 0$

This equation has no solution.

A a non-zero constant never equals zero.

Find the Vertex of $y = {x}^{2} -6x-11$

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,A${x}^{2}$+B$x$+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the $x$

coordinate is  3.0000 

Plugging into the parabola formula   3.0000  for  x  we can calculate the  

$y$ -coordinate :   $y$ = 1.0 * 3.00 * 3.00 - 6.0 * 3.00 - 11.0 or   $y$

= -20.000

Root plot for $y = {x}^{2} - 6x - 11$

Axis of Symmetry (dashed)  {x}={ 3.00} 

Vertex at  {x,y} = { 3.00,-20.00} 

 x -Intercepts (Roots) :

Root 1 at  {x,y} = {-1.47, 0.00} 

Root 2 at  {x,y} = { 7.47, 0.00}

(Please click above graph)

Solve

${x}^{2} \times - 6x = 11$

 by Completing The Square .

 Add  11  to both side of the equation :

  ${x}^{2} -6x = 11$

Now the clever bit: Take the coefficient of  x , which is  6 , divide by two, giving  3 , and finally square it giving  9 

Add  9  to both sides of the equation :

  On the right hand side we have :

   11  +  9    or,  (11/1)+(9/1) 

  The common denominator of the two fractions is  1   Adding  (11/1)+(9/1)  gives  20/1 

  So adding to both sides we finally get :

   x2-6x+9 = 20

Adding  9  has completed the left hand side into a perfect square :

${x}^{2} - 6x + 9 \\ (x - 3) • (x - 3) \\ {(x - 3)}^{2}$

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

${x}^{2} - 6x + 9 = 20 \: and \: {x}^{2} - 6x + 9 = {(x - 3)}^{2}$

then, according to the law of transitivity,

  

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

Note that the square root of

$= (x - 3)^{2} \:is \: \\ = {(x - 3)}^{ \frac{2}{2}} \\ = {(x - 3)}^{1} \\ = (x - 3)$

Now, applying the Square Root Principle to  Eq. #3.3.1  we get:

$x - 3 = \sqrt{20}$

Add  3  to both sides to obtain:

  

$x = 3 + \sqrt{20}$

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

${x}^{2} - 6x - 11 = 0$

   has two solutions:

$x = 3 + \sqrt{20} \\ or \: x = 3 - \sqrt{20}$

Solving   

${x}^{2} - 6x - 11$

by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  A${x}^{2}$+B$x$+C$= 0$, where  A, B  and  C  are numbers, often called coefficients, is given by :

$x = \frac{ - B± \sqrt{ {B}^{2 - 4AC} } }{2A}$

In our case,  A   =     1

                      B   =    -6

                      C   =  -11

Accordingly,  B2  -  4AC   

 = 36 - (-44)

               =  80

Applying the quadratic formula :

$x = \frac{6± \sqrt{80} }{2}$

Can  $\sqrt{80}$

be simplified ?

Yes! The prime factorization of  80   is

   2•2•2•2•5 

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).

√ 80   =  √ 2•2•2•2•5  

 = 2•2•√ 5 

           = ±  4 • √ 5

$\sqrt{80} = \sqrt{2•2•2•2•5} \\ = \sqrt[2•2•]{5} \\ = \sqrt[± 4•]{5}$

$\sqrt{5}$ rounded to 4 decimal digits, is   2.2361

 So now we are looking at:

           x  =  ( 6 ± 4 •  2.236 ) / 2

$x = \frac{ (6 ± 4 • 2.236)}{2}$

Two real solutions:

$x = \frac{(6+√80)}{2} =3+2 \sqrt{5} = 7.472 or x = \frac{6 - \sqrt{80} }{2} =3-2\sqrt{5} = -1.472$