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artcher [175]
3 years ago
9

Help find zeros for 9 and 10

Mathematics
1 answer:
Bingel [31]3 years ago
5 0
<span><span> x4-10x2+9=0</span> </span>Four solutions were found :<span> x = 3 x = -3 x = 1 x = -1</span>

Step by step solution :<span>Step  1  :</span>Skip Ad
<span>Equation at the end of step  1  :</span><span> ((x4) - (2•5x2)) + 9 = 0 </span><span>Step  2  :</span>Trying to factor by splitting the middle term

<span> 2.1 </span>    Factoring <span> x4-10x2+9</span> 

The first term is, <span> <span>x4</span> </span> its coefficient is <span> 1 </span>.
The middle term is, <span> <span>-10x2</span> </span> its coefficient is <span> -10 </span>.
The last term, "the constant", is  <span> +9 </span>

Step-1 : Multiply the coefficient of the first term by the constant <span> <span> 1</span> • 9 = 9</span> 

Step-2 : Find two factors of   9  whose sum equals the coefficient of the middle term, which is  <span> -10 </span>.

<span>     -9   +   -1   =   -10   That's it</span>


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -9  and  -1 
                     <span>x4 - 9x2</span> - <span>1x2 - 9</span>

Step-4 : Add up the first 2 terms, pulling out like factors :
                    <span>x2 • (x2-9)</span>
              Add up the last 2 terms, pulling out common factors :
                     1 • <span>(x2-9)</span>
Step-5 : Add up the four terms of step 4 :
                    <span>(x2-1)  •  (x2-9)</span>
             Which is the desired factorization

<span>Trying to factor as a Difference of Squares : </span>

<span> 2.2 </span>     Factoring: <span> x2-1</span> 

Theory : A difference of two perfect squares, <span> A2 - B2  </span>can be factored into <span> (A+B) • (A-B)

</span>Proof :<span>  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
         A2 <span>- AB + AB </span>- B2 = 
        <span> A2 - B2</span>

</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication. 

Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.

Check : 1 is the square of 1
Check : <span> x2  </span>is the square of  <span> x1 </span>

Factorization is :       (x + 1)  •  (x - 1) 

<span>Trying to factor as a Difference of Squares : </span>

<span> 2.3 </span>     Factoring: <span> x2 - 9</span> 

Check : 9 is the square of 3
Check : <span> x2  </span>is the square of  <span> x1 </span>

Factorization is :       (x + 3)  •  (x - 3) 

<span>Equation at the end of step  2  :</span> (x + 1) • (x - 1) • (x + 3) • (x - 3) = 0 <span>Step  3  :</span>Theory - Roots of a product :

<span> 3.1 </span>   A product of several terms equals zero.<span> 

 </span>When a product of two or more terms equals zero, then at least one of the terms must be zero.<span> 

 </span>We shall now solve each term = 0 separately<span> 

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

 </span>Any solution of term = 0 solves product = 0 as well.

<span>Solving a Single Variable Equation : </span>

<span> 3.2 </span>     Solve  :    x+1 = 0<span> 

 </span>Subtract  1  from both sides of the equation :<span> 
 </span>                     x = -1 

<span>Solving a Single Variable Equation : </span>

<span> 3.3 </span>     Solve  :    x-1 = 0<span> 

 </span>Add  1  to both sides of the equation :<span> 
 </span>                     x = 1 

<span>Solving a Single Variable Equation : </span>

<span> 3.4 </span>     Solve  :    x+3 = 0<span> 

 </span>Subtract  3  from both sides of the equation :<span> 
 </span>                     x = -3 

<span>Solving a Single Variable Equation : </span>

<span> 3.5 </span>     Solve  :    x-3 = 0<span> 

 </span>Add  3  to both sides of the equation :<span> 
 </span>                     x = 3 

Supplement : Solving Quadratic Equation Directly<span>Solving <span> x4-10x2+9</span>  = 0 directly </span>

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 

<span>Solving a Single Variable Equation : </span>

Equations which are reducible to quadratic :

<span> 4.1 </span>    Solve  <span> x4-10x2+9 = 0</span>

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  w , such that <span> w = x2</span>  transforms the equation into :
<span> w2-10w+9 = 0</span>

Solving this new equation using the quadratic formula we get two real solutions :
   9.0000  or   1.0000

Now that we know the value(s) of <span> w</span> , we can calculate <span> x</span>  since <span> x</span> <span> is  </span><span> √<span> w </span></span> 

Doing just this we discover that the solutions of 
  <span> x4-10x2+9 = 0</span>
  are either : 
  x =√<span> 9.000 </span>= 3.00000  or :
  x =√<span> 9.000 </span>= -3.00000  or :
  x =√<span> 1.000 </span>= 1.00000  or :
  x =√<span> 1.000 </span>= -1.00000 

Four solutions were found :<span> x = 3 x = -3 x = 1 x = -1</span>

<span>
Processing ends successfully</span>

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