Question 1186256
your equation is 2x^8 - 7x^4 - 3 = 0.


this is equivalent to:


2 * (x^4) ^ 2 - 7 * (x^4) ^ 1 - 3 = 0


if you replace x^4 with y, the equation becomes:


2y^2 - 7y - 3 = 0


this in standard form of ax^2 + bx + c = 0


when you replace x with y, it becomes:


ay^2 + by + c = 0


if you solve the quadratic equation using the quadratic formula, you will get:


y = 3.886000936 or y = -0.386000936


since y = x^4, solve this for x to get:


x = (3.886000936) ^ (1/4) or y = (-0.386000936) ^ (1/4).


you will find that the first option will get you x = plus or minus 1.404027859.


you will find that the second option will get you x = undefined because you can't take the even root of a negative number and get a real number as an answer.


you can get a complex number as an answer, but not a real number.


your solution becomes:


x = plus or minus fourth root of 3.886000936 which is equal to plus or minus 1.404027859.


if you graph the equation of y = 2x^8 - 7x^4 - 3, you will find that the roots are plus or minus 1.404 when rounded to 3 decimal places, as shown below.


<img src = "http://theo.x10hosting.com/2021/101711.jpg" >


here is the quadratic formula.


<pre>


               (-b plus or minus sqrt(b^2 - 4ac)
x =          --------------------------------------
                              2a

</pre>