# Solve for a 9/(a^2)+16/(a^2-25)=1 9a2+16a2-25=1
Factor each term.
Rewrite 25 as 52.
9a2+16a2-52=1
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=a and b=5.
9a2+16(a+5)(a-5)=1
9a2+16(a+5)(a-5)=1
Find the LCD of the terms in the equation.
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
a2,(a+5)(a-5),1
Since a2,(a+5)(a-5),1 contain both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.
Steps to find the LCM for a2,(a+5)(a-5),1 are:
1. Find the LCM for the numeric part 1,1,1.
2. Find the LCM for the variable part a2.
3. Find the LCM for the compound variable part a+5,a-5.
4. Multiply each LCM together.
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 factors for a2 are a⋅a, which is a multiplied by each other 2 times.
a2=a⋅a
a occurs 2 times.
The LCM of a2 is the result of multiplying all prime factors the greatest number of times they occur in either term.
a⋅a
Multiply a by a.
a2
The factor for a+5 is a+5 itself.
(a+5)=a+5
(a+5) occurs 1 time.
The factor for a-5 is a-5 itself.
(a-5)=a-5
(a-5) occurs 1 time.
The LCM of a+5,a-5 is the result of multiplying all factors the greatest number of times they occur in either term.
(a+5)(a-5)
The Least Common Multiple LCM of some numbers is the smallest number that the numbers are factors of.
a2(a+5)(a-5)
a2(a+5)(a-5)
Multiply each term by a2(a+5)(a-5) and simplify.
Multiply each term in 9a2+16(a+5)(a-5)=1 by a2(a+5)(a-5) in order to remove all the denominators from the equation.
9a2⋅(a2(a+5)(a-5))+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Simplify 9a2⋅(a2(a+5)(a-5))+16(a+5)(a-5)⋅(a2(a+5)(a-5)).
Simplify each term.
Cancel the common factor of a2.
Factor a2 out of a2(a+5)(a-5).
9a2⋅(a2((a+5)(a-5)))+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Cancel the common factor.
9a2⋅(a2((a+5)(a-5)))+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Rewrite the expression.
9⋅((a+5)(a-5))+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
9⋅((a+5)(a-5))+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Expand (a+5)(a-5) using the FOIL Method.
Apply the distributive property.
9⋅(a(a-5)+5(a-5))+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Apply the distributive property.
9⋅(a⋅a+a⋅-5+5(a-5))+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Apply the distributive property.
9⋅(a⋅a+a⋅-5+5a+5⋅-5)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
9⋅(a⋅a+a⋅-5+5a+5⋅-5)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Combine the opposite terms in a⋅a+a⋅-5+5a+5⋅-5.
Reorder the factors in the terms a⋅-5 and 5a.
9⋅(a⋅a-5a+5a+5⋅-5)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
9⋅(a⋅a+0+5⋅-5)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
9⋅(a⋅a+5⋅-5)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
9⋅(a⋅a+5⋅-5)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Simplify each term.
Multiply a by a.
9⋅(a2+5⋅-5)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Multiply 5 by -5.
9⋅(a2-25)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
9⋅(a2-25)+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Apply the distributive property.
9a2+9⋅-25+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Multiply 9 by -25.
9a2-225+16(a+5)(a-5)⋅(a2(a+5)(a-5))=1⋅(a2(a+5)(a-5))
Cancel the common factor of (a+5)(a-5).
Factor (a+5)(a-5) out of a2(a+5)(a-5).
9a2-225+16(a+5)(a-5)⋅((a+5)(a-5)(a2))=1⋅(a2(a+5)(a-5))
Cancel the common factor.
9a2-225+16(a+5)(a-5)⋅((a+5)(a-5)a2)=1⋅(a2(a+5)(a-5))
Rewrite the expression.
9a2-225+16⋅a2=1⋅(a2(a+5)(a-5))
9a2-225+16a2=1⋅(a2(a+5)(a-5))
9a2-225+16a2=1⋅(a2(a+5)(a-5))
25a2-225=1⋅(a2(a+5)(a-5))
25a2-225=1⋅(a2(a+5)(a-5))
Simplify 1⋅(a2(a+5)(a-5)).
Multiply a2(a+5)(a-5) by 1.
25a2-225=a2(a+5)(a-5)
Apply the distributive property.
25a2-225=(a2a+a2⋅5)(a-5)
Multiply a2 by a by adding the exponents.
Multiply a2 by a.
Raise a to the power of 1.
25a2-225=(a2a1+a2⋅5)(a-5)
Use the power rule aman=am+n to combine exponents.
25a2-225=(a2+1+a2⋅5)(a-5)
25a2-225=(a2+1+a2⋅5)(a-5)
25a2-225=(a3+a2⋅5)(a-5)
25a2-225=(a3+a2⋅5)(a-5)
Move 5 to the left of a2.
25a2-225=(a3+5⋅a2)(a-5)
Expand (a3+5a2)(a-5) using the FOIL Method.
Apply the distributive property.
25a2-225=a3(a-5)+5a2(a-5)
Apply the distributive property.
25a2-225=a3a+a3⋅-5+5a2(a-5)
Apply the distributive property.
25a2-225=a3a+a3⋅-5+5a2a+5a2⋅-5
25a2-225=a3a+a3⋅-5+5a2a+5a2⋅-5
Simplify and combine like terms.
Simplify each term.
Multiply a3 by a by adding the exponents.
Multiply a3 by a.
Raise a to the power of 1.
25a2-225=a3a1+a3⋅-5+5a2a+5a2⋅-5
Use the power rule aman=am+n to combine exponents.
25a2-225=a3+1+a3⋅-5+5a2a+5a2⋅-5
25a2-225=a3+1+a3⋅-5+5a2a+5a2⋅-5
25a2-225=a4+a3⋅-5+5a2a+5a2⋅-5
25a2-225=a4+a3⋅-5+5a2a+5a2⋅-5
Move -5 to the left of a3.
25a2-225=a4-5⋅a3+5a2a+5a2⋅-5
Multiply a2 by a by adding the exponents.
Move a.
25a2-225=a4-5a3+5(a⋅a2)+5a2⋅-5
Multiply a by a2.
Raise a to the power of 1.
25a2-225=a4-5a3+5(a1a2)+5a2⋅-5
Use the power rule aman=am+n to combine exponents.
25a2-225=a4-5a3+5a1+2+5a2⋅-5
25a2-225=a4-5a3+5a1+2+5a2⋅-5
25a2-225=a4-5a3+5a3+5a2⋅-5
25a2-225=a4-5a3+5a3+5a2⋅-5
Multiply -5 by 5.
25a2-225=a4-5a3+5a3-25a2
25a2-225=a4-5a3+5a3-25a2
25a2-225=a4+0-25a2
25a2-225=a4-25a2
25a2-225=a4-25a2
25a2-225=a4-25a2
25a2-225=a4-25a2
Solve the equation.
Since a is on the right side of the equation, switch the sides so it is on the left side of the equation.
a4-25a2=25a2-225
Move all the expressions to the left side of the equation.
Move 25a2 to the left side of the equation by subtracting it from both sides.
a4-25a2-25a2=-225
Move 225 to the left side of the equation by adding it to both sides.
a4-25a2-25a2+225=0
a4-25a2-25a2+225=0
Subtract 25a2 from -25a2.
a4-50a2+225=0
Substitute u=a2 into the equation. This will make the quadratic formula easy to use.
u2-50u+225=0
u=a2
Factor u2-50u+225 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 225 and whose sum is -50.
-45,-5
Write the factored form using these integers.
(u-45)(u-5)=0
(u-45)(u-5)=0
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
u-45=0
u-5=0
Set the first factor equal to 0 and solve.
Set the first factor equal to 0.
u-45=0
Add 45 to both sides of the equation.
u=45
u=45
Set the next factor equal to 0 and solve.
Set the next factor equal to 0.
u-5=0
Add 5 to both sides of the equation.
u=5
u=5
The final solution is all the values that make (u-45)(u-5)=0 true.
u=45,5
Substitute the real value of u=a2 back into the solved equation.
a2=45
(a2)1=5
Solve the first equation for a.
a2=45
Solve the equation for a.
Take the square root of both sides of the equation to eliminate the exponent on the left side.
a=±45
The complete solution is the result of both the positive and negative portions of the solution.
Simplify the right side of the equation.
Rewrite 45 as 32⋅5.
Factor 9 out of 45.
a=±9(5)
Rewrite 9 as 32.
a=±32⋅5
a=±32⋅5
Pull terms out from under the radical.
a=±35
a=±35
The complete solution is the result of both the positive and negative portions of the solution.
First, use the positive value of the ± to find the first solution.
a=35
Next, use the negative value of the ± to find the second solution.
a=-35
The complete solution is the result of both the positive and negative portions of the solution.
a=35,-35
a=35,-35
a=35,-35
a=35,-35
Solve the second equation for a.
(a2)1=5
Solve the equation for a.
Take the 1th root of each side of the equation to set up the solution for a
(a2)1⋅11=51
Remove the perfect root factor a2 under the radical to solve for a.
a2=51
Take the square root of both sides of the equation to eliminate the exponent on the left side.
a=±51
The complete solution is the result of both the positive and negative portions of the solution.
Evaluate 51 as 5.
a=±5
The complete solution is the result of both the positive and negative portions of the solution.
First, use the positive value of the ± to find the first solution.
a=5
Next, use the negative value of the ± to find the second solution.
a=-5
The complete solution is the result of both the positive and negative portions of the solution.
a=5,-5
a=5,-5
a=5,-5
a=5,-5
The solution to a4-50a2+225=0 is a=35,-35,5,-5.
a=35,-35,5,-5
a=35,-35,5,-5
Exclude the solutions that do not make 9a2+16a2-25=1 true.
a=35,-35,5,-5
The result can be shown in multiple forms.
Exact Form:
a=35,-35,5,-5
Decimal Form:
a=6.70820393…,-6.70820393…,2.23606797…,-2.23606797…
Solve for a 9/(a^2)+16/(a^2-25)=1

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