can someone do this with steps please?

Question

solve each equation.

16x^4-41x^2+25=0
4x^-2+5x^-1+1=0

Answer 1

These equations are just quadratic equations in disguise. Let y=x^2. We can rewrite the first equation as 16y^2-41y+25=0 and solve by factoring.

(16y-25)(y-1)=0. Either 16y-25=0 or y-1=0, so y=25/16 or y=1. But we want x not y. However, y=x^2, so x is just the square root of y. Thus, the solutions to the equation are x= -5/4, -1, 1, 5/4. You can use a similar trick (y=x^-1) to solve the second equation, but I'll leave that to you.

Answer 2

Let's do the second one first so you will see a process:

  4x^-2  +  5x^-1  +  1  =  0

  4     +  5x   +  x²   = 0       [multiply all by x²]

  x²   +  5x    +   4   =  0         [re-arrange terms]

   (x + 4)(x + 1) = 0            [either factor or use quadratic formula]

     either (x+4)=0 or (x+1)=0           [the multiplicative property of zero]

     either   x=-4  or   x=-1

 

First one:

   16x⁴   -  41x²   +  25  =  0

   Let y=x²           [substitution is a process that makes problem-solving easier]

     16y²  -  41y   +  25  =  0

     (16y - 25)(y - 1) = 0          [either factor or use quadratic formula]

     either (16y-25)=0  or  (y-1)=0          [the multiplicative property of zero]

     Now, you may either solve for y then substitute x² or you may (this is easier) substitute x² first then solve for x.

 

  Remember these two processes.