Just square the answer

$x+\frac 1x = 5$

$\Rightarrow (x+\frac 1x)^2 = 5^2$

$\Rightarrow x^2+\frac xx+\frac xx+ \frac 1{x^2} = 5^2$

$\Rightarrow x^2+2+ \frac 1{x^2} = 25$

$\Rightarrow x^2+\frac 1{x^2} = 25-2$

$\Rightarrow x^2+\frac 1{x^2} = 23$

let $x=a$ , $\frac{1}{x}=b$ ,

then $ab=1$

$a+b=5$

$(a+b)^2=5^2$

$a^2+b^2+2ab=25$

$a^2+b^2+2=25$

$a^2+b^2=25-2$

$a^2+b^2=23$

$x^2+\frac{1}{x^2}=23$

On squaring both sides and subtracting 2 from each side...you get the answer to be 23!!

Slightly Longer Than Other Answers But Makes More Sense? Rearrange equation to get quadratic. Solve quadratic for x and then put back into equation. Rearrange by moving x across and then multiplying by x X(5-X)=1 multiply out 5X-X^2=1 X^2-5X+1=0 SOLVE PUT X BACK IN

x+ 1/x = 5

squaring both sides

x^2 + 1/x^2 +2 = 25

x^2 + 1/x^2 = 23

$x+\frac{1}{x}=5 \implies (x+\frac{1}{x})^2=25 \implies x^2+2+\frac{1}{x^2}=25 \implies x^2+\frac{1}{x^2}=\boxed{\large{23}}$

we know that, (a^2+b^2)=(a+b)^2-2ab

so, in this case,x^2+1/x^2=(x+1/x)^2- 2 .x . 1/x

                                      =5^2-2=23

Solution 1

Working on $x +$ $\dfrac{1}{x}$ $= 5$

$x +$ $\dfrac{1}{x}$ $= 5$

$(x + \dfrac{1}{x})^2$ $= 5^2$

We know that, $(a + b)^2 = a^2 + 2ab + b^2$ so,

$x^2 + 2 \cdot x$ $\dfrac{1}{x}$ $+$ $(\dfrac{1}{x})^2$ $= 25$

$x^2 + 2 +$ $\dfrac{1}{x^2}$ $= 25$ ;[ $x$ $\dfrac{1}{x}$ is cancelled]

$x^2 +$ $\dfrac{1}{x^2}$ $= 25 - 2$

Therefore $x^2 +$ $\dfrac{1}{x^2}$ $= \boxed{23}$

Solution 2

Given that,

• $x +$ $\dfrac{1}{x}$ $= 5$

Now,

$(x +$ $\dfrac{1}{x})^2$ $- 2 \cdot x$ $\dfrac{1}{x}$

$(5)^2$ $- 2$ ;[ $x \dfrac{1}{x}$ is cancelled]

$= \boxed{23}$

Square both sides so we will get, X^2 +2x1/x+1/x^2=25 Thus. X2+1/x2=25-2=23

x2+1/x2 = (x+1/x)2-2= 25-2=23

Square both side . after Solve the answer is 23

(x+1\x)^2=25 X^2+2 +(1\x^2)=25 X^2+(1\x^2)=23

x^2+1/x^2 =(x+1/x)^2-2.x.1/x =(5)^2-2 =25-2 =23

Here's another way to solve the problem (For those of you who like calculators):

Step 1: Manipulate the first equation into a solvable form:

x + 1/x = 5 -> 1/x = 5 - x -> 1 =x(5-x) -> 1 = 5x - x^2 -> x^2 - 5x + 1 = 0

Step 2: Use the Quadratic Formula to arrive at these answers:

x=(5+sqrt 21)/2, x=(5-sqrt 21)/2

Step 3: Plug in either value of X into the second equation, x^2+1/x^2=?, and you should get 23.

Use solver in excel.

Set A1 to 1. Set A2 to =A1 + (1/A1) Set B2 = A1^2+(1/(A1^2)) Run solver add in, find solution for B1 by manipulating A1. cell B2 has your solution.

Solver has destroyed my paper and pencil algebra ability.

Since; x+1/x =5 Squaring on both the side (x+1/x)² = x²+1/x²+2•x•1/x= 5² Or ; x²+1/x²+2= 25 Or; x²+1/x²= 25-2 =23

On squaring both sides, you get the answer to be 23

See..do (x+1/x)^2.. Exapnd it normally and subtract 2 to get the answer...so according to.our logic... The solution is... (5)^2-2=23

x^2 + (i/x)^2 = (x + 1/x)^2 - 2 = 5^2 - 2 = 23

just use a formula( x+1 upon x) whole square = x square + 1 upon x square. hence, by putting value you will get the answer

square both sides and you're done!

Squaring both sides we get the value of expression in question 25 -2 = 23 Ans K.KL.GARG,India

Square both sides of the equation. After the first method, you will come up with an answer that expresses the arithmetical equation x(squared) + 2 + 1/x(squared) = 25. Simply put 2 to the right side of the equation. Therefore, when 2 is transposed, it will become -2.. There is 25 at the right side of the equation: 25-2=23. I feel apologetic for the solution. I know it's awfully cryptic to decipher well.

$(x+\frac{1}{x})^{2}$ = $x^{2}$ + $2$ + $\frac{1}{x^{2}}$

So,

$x^{2}$ + $\frac{1}{x^{2}}$ = $(x+\frac{1}{x})^{2}$ - $2$

$x^{2}$ + $\frac{1}{x^{2}}$ = $5^{2}$ - $2$

$x^{2}$ + $\frac{1}{x^{2}}$ = $25$ - $2$

$x^{2}$ + $\frac{1}{x^{2}}$ = $23$

x + 1/x =5 || (x+1/x)^2 = 5^2 || We know that (a + b)^2 = a^2 + 2ab + b^2 || continuance : x^2 + 2.x.1/x +(1/x)^2 = 25 || x^2 + 2 + 1/x^2 = 25 || x^2 + 1/x^2 = 25-2 || x^2 + 1/x^2 = 23

we actually have 2 solution for this. first undo the 1/x. this mean we have to multiply everything in the equation by x. we will have this x^2+1=5x. now rearrange the equation so it equal to zero. x^2-5x+1=0. now complete the square so we end up with (x-2.5)^2-6.25+1=0. tidy it up and we get (x-2.5)^2-5.2=0 . rearrange the equation again to give (x-2.5)^2=5.25 . this mean we have x-2.5=+ or - square root of 5.25 (which is +-2.29). now the equation is x-2.5=+-2.29 now just take the rearrange to make x= -0.4 or 23

as we know a square + b square = a+b sqaure -2ab, applying this property we get (x + 1/x)square -2 * x* 1/x=25-2=23

squaring both sides. (X+1/X)^2 = (5)^2 ; X^2 + (1/X^2) + 2 = 25 ; X^2 + (1/X^2) = 23

5 x 5 -2 x X 1/X=23