Question

If x2-x-1=0, then find the value of x3–2x+1​

Answer 1

Answer:

First, the brute force method. x2−x−1=0 has two solutions:

x=1±5√2

These can be calculated from the quadratic equation.

Finally, plugging either value into x3−2x+1 gives the answer as 2.

A slightly more clever method is to use the first equation to reduce the order of the second equation. Specifically we apply this form of the first equation:

x2=x+1

To get:

x3−2x+1=x(x+1)−2x+1

=x2+x−2x+1

=(x+1)−x+1

=2

I hope this is helpful for you

Answer 2

Your answer is Here mate

If x2−x−1=0 , then what is the value of x3−2x+1?

If x2−x−1=0 , then what is the value of x3−2x+1? There’s a very quick answer to this one if you're familiar with the golden ratio and its conjugate together with their properties.

The relevant property is that the ratios each satisfy xn=xn−1+xn−2 for any power n. Firstly this means that they satisfy:

x2=x+1 and so they are the two roots of the first given equation.

The same property also means that they satisfy:x3=x2+x

and inserting this into the second given equation converts it to:x2−x+1=(x2−x−1)+2=2

In other words the value of the second given equation must equal 2 for both roots of the first

Hope it helps you