Which quadratic equation is equivalent to [tex]u^2-11 u+24=0[/tex] where [tex]u=(x^2-1)[/tex]?
[tex](u^2)^2-11(u^2)+24[/tex] where [tex]u=(x^2-1)[/tex]
[tex]u^2+1-11 u+24=0[/tex] where [tex]u=(x^2-1)[/tex]
[tex](u^2-1)^2-11(u^2-1)+24[/tex] where [tex]u=(x^2-1)[/tex]
Answer (2)
Substitute u = x 2 − 1 into the original equation u 2 − 11 u + 24 = 0 .
Expand the resulting equation to get x 4 − 13 x 2 + 36 = 0 .
Compare the expanded form with the expanded forms of the given options.
The equivalent equation is ( x 2 − 1 ) 2 − 11 ( x 2 − 1 ) + 24 = 0 , which simplifies to x 4 − 13 x 2 + 36 = 0 . Therefore, the equivalent equation is ( x 2 − 1 ) 2 − 11 ( x 2 − 1 ) + 24 = 0 .
Explanation
Understanding the Problem We are given the quadratic equation u 2 − 11 u + 24 = 0 where u = x 2 − 1 . We need to find an equivalent equation in terms of x from the given options.
Substitution First, let's substitute u = x 2 − 1 into the original equation: ( x 2 − 1 ) 2 − 11 ( x 2 − 1 ) + 24 = 0
Expansion Now, let's expand this equation: ( x 4 − 2 x 2 + 1 ) − 11 x 2 + 11 + 24 = 0 x 4 − 2 x 2 + 1 − 11 x 2 + 11 + 24 = 0 x 4 − 13 x 2 + 36 = 0
Checking the options Now, let's check the given options by substituting u = x 2 − 1 into each and expanding.
Option 1: ( u 2 ) 2 − 11 ( u 2 ) + 24 = 0 where u = ( x 2 − 1 ) This becomes (( x 2 − 1 ) 2 ) 2 − 11 (( x 2 − 1 ) 2 ) + 24 = 0 , which expands to a degree 8 polynomial, so it's not equivalent.
Option 2: u 2 + 1 − 11 u + 24 = 0 where u = ( x 2 − 1 ) This becomes ( x 2 − 1 ) 2 + 1 − 11 ( x 2 − 1 ) + 24 = 0 , which expands to x 4 − 2 x 2 + 1 + 1 − 11 x 2 + 11 + 24 = x 4 − 13 x 2 + 37 = 0 , so it's not equivalent.
Option 3: ( u 2 − 1 ) 2 − 11 ( u 2 − 1 ) + 24 where u = ( x 2 − 1 ) This is not a valid option since it contains u 2 − 1 inside the parenthesis, but u = x 2 − 1 , so it should be ( x 2 − 1 ) 2 − 11 ( x 2 − 1 ) + 24 = 0 , which is the original equation.
Expanding the correct option Let's consider the equation ( x 2 − 1 ) 2 − 11 ( x 2 − 1 ) + 24 = 0 . Expanding this, we get: ( x 4 − 2 x 2 + 1 ) − 11 ( x 2 − 1 ) + 24 = 0 x 4 − 2 x 2 + 1 − 11 x 2 + 11 + 24 = 0 x 4 − 13 x 2 + 36 = 0
Conclusion The equation x 4 − 13 x 2 + 36 = 0 is equivalent to ( x 2 − 1 ) 2 − 11 ( x 2 − 1 ) + 24 = 0 . Therefore, the correct option is the one that simplifies to this equation. The option that is equivalent to the original equation is ( x 2 − 1 ) 2 − 11 ( x 2 − 1 ) + 24 = 0 , which can be written as ( u ) 2 − 11 ( u ) + 24 = 0 where u = x 2 − 1 .
Examples
Imagine you are designing a bridge and need to ensure that the forces acting on it are balanced. By using quadratic equations and variable substitution, you can model the forces and find the points where the bridge is most stable. This helps engineers design safer and more efficient structures. Similarly, in physics, when analyzing projectile motion, substituting variables in quadratic equations can simplify the calculations and help determine the trajectory of an object.
The correct quadratic equation equivalent to u 2 − 11 u + 24 = 0 when substituting u = x 2 − 1 is ( x 2 − 1 ) 2 − 11 ( x 2 − 1 ) + 24 = 0 . This option corresponds to the original equation upon expansion. Thus, the answer is option 3.
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