If x1 and x2 are two solutions of the equations(log2x)^3 – 3log2x + 5 = 0, then the product x1x2 is?

Question

If x1 and x2 are two solutions of the equations(log2x)^3 – 3log2x + 5 = 0, then the product x1x2 is
"give calculation process"

Answer 1

The equation is a quadratic in log(2x). We do not need to solve this quadratic to know that the sum of the two roots
log(2x(1)) and log(2x(2)) is 3. This is because if the quadratic equation ax^2 + bx + c = 0 has roots r and s then r + s = -b/a.
log(2x(1)) + log(2x(2)) = 3
log(4x(1)*x(2)) = 3
log(4) + log(x(1)*x(2)) = 3
log(x(1)*x(2)) = 3 - log(4)
x(1)*x(2) = 10^(3 - log(4))
assuming that the logs in the question were to base 10.

Edit. Just noticed that the original equation is not a quadratic but a cubic. Was this intended? Since only two solutions are mentioned in the question then I suspect that it was meant to be a quadratic but I might be going totally in the wrong direction.

Answer 2

^ <-- that is the expontent symbol, it basically symnolizes that something is to the power of what ever number ^2 = squared or to the power of 2. They used the Foil Method (First, Outter, Inner, Last) x*x= x^2 x*1=x ( you don't need the 1 it is understood automatically) 5*x= 5x 5*1=5 Now you combine like terms x^2+x+5x+5= x^2 + 6x+5 now you just set it equal to 0 which is the same. you basically break down the first formula. You don't need the =0 part you can put that in at the end, it will still be the same. Keep in mind that x times x is always x squared (x ^2) x^2 + 6x + 5 (x+_) (x+_) what two numbers multiplied together give u (positive) 5 and added together give you 6? The only option is 5 and 1. Now since they are both positive it doesn't matter where they go. This problem you have is basically factoring trinomials.

Answer 3

Hai,I think your question is wrong because
you have told that x1 and x2 are two solutions of the above equation but it is not a quadratic equation rather it is a cubic equation therfore there are 3 roots for the above equation,therefore we cannot find the product of them because we dont know which root u have asked...
I hope u have understood my answer....
Question might be (log2x)^2-3(log2x)+5=0
then product of the roots=(c/a) if it is of the form ax^2+bx+c=0
Therefore product of the roots = (5/(log2x))

Answer 4

call z=log2x
z^3-3z+5=y
Let´ s see the number of roots
y´=3z^2-3 =0 so z=+-1 and the sign is +++-1----1+++++
y(-1)>0 and y(1)>0 so there is only one real root less than -1
so if the product z1*z2 has to be real you have to take the complex conjugates roots.
I don´t think that this was the intention of the problem because you can find all the roots only by appoximation