The sum of a number and its reciprocal is 2, find the numbe

[ginny1029]

so i think is should be

x + 1/x = 2

is that right

and if so would i then multiple everything by x to get rid of the fraction such as

x (x) + 1/x (x) = 2(x)

so x^2 = 2x

and again is so then what?

 

[Loren]

so i think is should be

x + 1/x = 2

is that right

and if so would i then multiple everything by x to get rid of the fraction such as

x (x) + 1/x (x) = 2(x) <<< Good to here which means \(\displaystyle x\cdot x + \frac{1}{x}\cdot \frac{x}{1} = 2x\). What is the simplification of this?

so x^2 = 2x

and again is so then what?

 

[ginny1029]

i think it is x^2 =2x

is that right?

to get x i would divide both sides 2

x^2 / 2 = 2x / 2

x^2 / 2 = x

 

[Loren]

i think it is x^2 =2x <<<Wrong. What happened when you evaluated (1/x)(x)? Does it just disappear?

is that right?

to get x i would divide both sides 2

x^2 / 2 = 2x / 2

x^2 / 2 = x

 

[ginny1029]

so is it x^2 + 1 = 2x?

 

[Mrspi]

so is it x^2 + 1 = 2x?

Yes, it is.

Now...get one side equal to 0:

x^2 - 2x + 1 = 0

You will have to use the quadratic formula to solve this, since the expression on the left side does not factor.

 

[ginny1029]

ok so from the beginning here is what I did

2 = x + (1/x)
2(x) = x (x) + (1/x)(x)
2x = x^2 + 1
0 = x^2 - 2x + 1
0 = (x-1) (x-1)
0 + x-1
0 + 1 = x
1 = x

is that about right?

Thanks for all the help

 

[Mrspi]

ok so from the beginning here is what I did

2 = x + (1/x)
2(x) = x (x) + (1/x)(x)
2x = x^2 + 1
0 = x^2 - 2x + 1
0 = (x-1) (x-1)
0 + x-1
0 + 1 = x
1 = x

is that about right?

Thanks for all the help

You can check to see if your answer is correct. You got x = 1. Substitute 1 for x in the original equation, and see if you get a true statement.