Find the best solutions to your problems with the help of IDNLearn.com's experts. Join our interactive community and get comprehensive, reliable answers to all your questions.
Sagot :
[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)}.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.
