Answer :
[tex]16{ x }^{ 4 }-41{ x }^{ 2 }+25=0[/tex]
[tex]{ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0[/tex]
First of all to make our equation simpler, we'll equal [tex] x^{2} [/tex] to a variable like 'a'.
So,
[tex]{ x }^{ 2 }=a[/tex]
Now let's plug [tex] x^{2} [/tex] 's value (a) into the equation.
[tex]16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0[/tex]
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 (fourth-degree equation) )
Let's solve for a.
The formula used to solve quadratic equations ;
[tex]\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t } [/tex]
The formula is used in an equation formed like this :
[tex]t{ x }^{ 2 }+bx+c=0[/tex]
In our equation,
t=16 , b=-41 and c=25
Let's plug the values in the formula to solve.
[tex]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 } [/tex]
So the solution set :
[tex]\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\} [/tex]
We found a's value.
Remember,
[tex]{ x }^{ 2 }=a[/tex]
So after we found a's solution set, that means.
[tex]{ x }^{ 2 }=\frac { 50 }{ 32 } [/tex]
and
[tex]{ x }^{ 2 }=1[/tex]
We'll also solve this equations to find x's solution set :)
[tex]{ 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 } [/tex]
[tex]{ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1[/tex]
So the values x has are :
[tex]\frac { 5 }{ 4 } [/tex] , [tex]-\frac { 5 }{ 4 } [/tex] , [tex]1[/tex] and [tex]-1[/tex]
Solution set :
[tex]x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\} [/tex]
I hope this was clear enough. If not please ask :)
[tex]{ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0[/tex]
First of all to make our equation simpler, we'll equal [tex] x^{2} [/tex] to a variable like 'a'.
So,
[tex]{ x }^{ 2 }=a[/tex]
Now let's plug [tex] x^{2} [/tex] 's value (a) into the equation.
[tex]16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0[/tex]
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 (fourth-degree equation) )
Let's solve for a.
The formula used to solve quadratic equations ;
[tex]\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t } [/tex]
The formula is used in an equation formed like this :
[tex]t{ x }^{ 2 }+bx+c=0[/tex]
In our equation,
t=16 , b=-41 and c=25
Let's plug the values in the formula to solve.
[tex]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 } [/tex]
So the solution set :
[tex]\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\} [/tex]
We found a's value.
Remember,
[tex]{ x }^{ 2 }=a[/tex]
So after we found a's solution set, that means.
[tex]{ x }^{ 2 }=\frac { 50 }{ 32 } [/tex]
and
[tex]{ x }^{ 2 }=1[/tex]
We'll also solve this equations to find x's solution set :)
[tex]{ 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 } [/tex]
[tex]{ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1[/tex]
So the values x has are :
[tex]\frac { 5 }{ 4 } [/tex] , [tex]-\frac { 5 }{ 4 } [/tex] , [tex]1[/tex] and [tex]-1[/tex]
Solution set :
[tex]x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\} [/tex]
I hope this was clear enough. If not please ask :)
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
+ 5 + 5 - 5 - 5 - 1 - 1 + 1 + 1
4x = 5 4x = -5 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)}.
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
+ 5 + 5 - 5 - 5 - 1 - 1 + 1 + 1
4x = 5 4x = -5 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)}.
