100×2-25+1=2xx-5

Subtract 1 from both sides of the equation.

100×2-25=2xx-5-1

Rewrite 25 as 52.

100×2-52=2xx-5-1

Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=x and b=5.

100(x+5)(x-5)=2xx-5-1

100(x+5)(x-5)=2xx-5-1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

(x+5)(x-5),x-5,1

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number 1 is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of 1,1,1 is the result of multiplying all prime factors the greatest number of times they occur in either number.

1

The factor for x+5 is x+5 itself.

(x+5)=x+5

(x+5) occurs 1 time.

The factor for x-5 is x-5 itself.

(x-5)=x-5

(x-5) occurs 1 time.

The LCM of x+5,x-5,x-5 is the result of multiplying all factors the greatest number of times they occur in either term.

(x+5)(x-5)

(x+5)(x-5)

Multiply each term in 100(x+5)(x-5)=2xx-5-1 by (x+5)(x-5) in order to remove all the denominators from the equation.

100(x+5)(x-5)⋅((x+5)(x-5))=2xx-5⋅((x+5)(x-5))-1⋅((x+5)(x-5))

Cancel the common factor of (x+5)(x-5).

Cancel the common factor.

100(x+5)(x-5)⋅((x+5)(x-5))=2xx-5⋅((x+5)(x-5))-1⋅((x+5)(x-5))

Rewrite the expression.

100=2xx-5⋅((x+5)(x-5))-1⋅((x+5)(x-5))

100=2xx-5⋅((x+5)(x-5))-1⋅((x+5)(x-5))

Simplify 2xx-5⋅((x+5)(x-5))-1⋅((x+5)(x-5)).

Simplify each term.

Cancel the common factor of x-5.

Factor x-5 out of (x+5)(x-5).

100=2xx-5⋅((x-5)(x+5))-1⋅((x+5)(x-5))

Cancel the common factor.

100=2xx-5⋅((x-5)(x+5))-1⋅((x+5)(x-5))

Rewrite the expression.

100=2x⋅(x+5)-1⋅((x+5)(x-5))

100=2x⋅(x+5)-1⋅((x+5)(x-5))

Apply the distributive property.

100=2x⋅x+2x⋅5-1⋅((x+5)(x-5))

Multiply x by x by adding the exponents.

Move x.

100=2(x⋅x)+2x⋅5-1⋅((x+5)(x-5))

Multiply x by x.

100=2×2+2x⋅5-1⋅((x+5)(x-5))

100=2×2+2x⋅5-1⋅((x+5)(x-5))

Multiply 5 by 2.

100=2×2+10x-1⋅((x+5)(x-5))

Expand (x+5)(x-5) using the FOIL Method.

Apply the distributive property.

100=2×2+10x-1⋅(x(x-5)+5(x-5))

Apply the distributive property.

100=2×2+10x-1⋅(x⋅x+x⋅-5+5(x-5))

Apply the distributive property.

100=2×2+10x-1⋅(x⋅x+x⋅-5+5x+5⋅-5)

100=2×2+10x-1⋅(x⋅x+x⋅-5+5x+5⋅-5)

Combine the opposite terms in x⋅x+x⋅-5+5x+5⋅-5.

Reorder the factors in the terms x⋅-5 and 5x.

100=2×2+10x-1⋅(x⋅x-5x+5x+5⋅-5)

Add -5x and 5x.

100=2×2+10x-1⋅(x⋅x+0+5⋅-5)

Add x⋅x and 0.

100=2×2+10x-1⋅(x⋅x+5⋅-5)

100=2×2+10x-1⋅(x⋅x+5⋅-5)

Simplify each term.

Multiply x by x.

100=2×2+10x-1⋅(x2+5⋅-5)

Multiply 5 by -5.

100=2×2+10x-1⋅(x2-25)

100=2×2+10x-1⋅(x2-25)

Apply the distributive property.

100=2×2+10x-1×2-1⋅-25

Rewrite -1×2 as -x2.

100=2×2+10x-x2-1⋅-25

Multiply -1 by -25.

100=2×2+10x-x2+25

100=2×2+10x-x2+25

Subtract x2 from 2×2.

100=x2+10x+25

100=x2+10x+25

100=x2+10x+25

Rewrite the equation as x2+10x+25=100.

x2+10x+25=100

Move 100 to the left side of the equation by subtracting it from both sides.

x2+10x+25-100=0

Subtract 100 from 25.

x2+10x-75=0

Factor x2+10x-75 using the AC method.

Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is -75 and whose sum is 10.

-5,15

Write the factored form using these integers.

(x-5)(x+15)=0

(x-5)(x+15)=0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

x-5=0

x+15=0

Set the first factor equal to 0 and solve.

Set the first factor equal to 0.

x-5=0

Add 5 to both sides of the equation.

x=5

x=5

Set the next factor equal to 0 and solve.

Set the next factor equal to 0.

x+15=0

Subtract 15 from both sides of the equation.

x=-15

x=-15

The final solution is all the values that make (x-5)(x+15)=0 true.

x=5,-15

x=5,-15

Exclude the solutions that do not make 100×2-25+1=2xx-5 true.

x=-15

Solve for x 100/(x^2-25)+1=(2x)/(x-5)