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Soloha48 [4]
3 years ago
6

How would I solve 16x^4-41x^2+25=0 ???

Mathematics
2 answers:
hichkok12 [17]3 years ago
8 0
16x⁴ - 41x² + 25 = 0
16x⁴ - 16x² - 25x² + 25 = 0
16x²(x²) - 16x²(1) - 25(x²) + 25(1) = 0
16(x² - 1) - 25(x² - 1) = 0
(16x² - 25)(x² - 1) = 0
(16x² + 20x - 20x - 25)(x² - x + x - 1) = 0
(4x(4x) + 4x(5) - 5(4x) - 5(5))(x(x) - x(1) + 1(x) - 1(1)) = 0
(4x(4x + 5) - 5(4x + 5))(x(x - 1) + 1(x - 1)) = 0
(4x - 5)(4x + 5)(x + 1)(x - 1) = 0
4x - 5 = 0  or  4x + 5 = 0  or  x + 1 = 0  or  x - 1 = 0
<u>    + 5 + 5</u>       <u>      - 5  - 5</u>      <u>    - 1  - 1</u>       <u>   + 1 + 1</u>
     <u>4x</u> = <u>5</u>              <u>4x</u> = <u>-5</u>          x = -1             x = 1
      4     4               4      4
       x = 1¹/₄            x = -1¹/₄

The solution set is equal to {(1¹/₄, -1¹/₄, -1, 1)}.
sveticcg [70]3 years ago
3 0
16{ x }^{ 4 }-41{ x }^{ 2 }+25=0

{ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0



First of all to make our equation simpler, we'll equal x^{2} to a variable like 'a'.

So,

{ x }^{ 2 }=a

Now let's plug x^{2} 's value (a) into the equation.

16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (<span>fourth-degree <span>equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c }  }{ 2\cdot t }

The formula is used in an equation formed like this :
</span></span>
t{ x }^{ 2 }+bx+c=0

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 }  }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 }  }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 }  }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }

So the solution set :

\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}

We found a's value.

Remember,

{ x }^{ 2 }=a

So after we found a's solution set, that means.

{ x }^{ 2 }=\frac { 50 }{ 32 }

and

{ x }^{ 2 }=1

We'll also solve this equations to find x's solution set :)

{ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 }  } \\ \\ x=\quad \pm \frac { 5 }{ 4 }

{ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1

So the values x has are :

\frac { 5 }{ 4 } , -\frac { 5 }{ 4 } , 1 and -1

Solution set :

x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}

I hope this was clear enough. If not please ask :)



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