Solve for a 9/(a^2)+16/(a^2-25)=1

$\frac{9}{a^{2}} + \frac{16}{a^{2} - 25} = 1$

Step 1

Factor each term.

Rewrite $25$ as $5^{2}$ .

$\frac{9}{a^{2}} + \frac{16}{a^{2} - 5^{2}} = 1$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = a$ and $b = 5$ .

$\frac{9}{a^{2}} + \frac{16}{(a + 5) (a - 5)} = 1$

Step 2

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.

$a^{2}, (a + 5) (a - 5), 1$

Since $a^{2}, (a + 5) (a - 5), 1$ contains 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 $a^{2}, (a + 5) (a - 5), 1$ are:

1. Find the LCM for the numeric part $1, 1, 1$ .

2. Find the LCM for the variable part $a^{2}$ .

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 $a^{2}$ are $a \cdot a$ , which is $a$ multiplied by each other $2$ times.

$a^{2} = a \cdot a$

$a$ occurs $2$ times.

The LCM of $a^{2}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$a \cdot a$

Multiply $a$ by $a$ .

$a^{2}$

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.

$a^{2} (a + 5) (a - 5)$

Step 3

Multiply each term in $\frac{9}{a^{2}} + \frac{16}{(a + 5) (a - 5)} = 1$ by $a^{2} (a + 5) (a - 5)$ to eliminate the fractions.

Multiply each term in $\frac{9}{a^{2}} + \frac{16}{(a + 5) (a - 5)} = 1$ by $a^{2} (a + 5) (a - 5)$ .

$\frac{9}{a^{2}} (a^{2} (a + 5) (a - 5)) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Simplify the left side.

Simplify each term.

Cancel the common factor of $a^{2}$ .

Factor $a^{2}$ out of $a^{2} (a + 5) (a - 5)$ .

$\frac{9}{a^{2}} (a^{2} ((a + 5) (a - 5))) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Cancel the common factor.

$\frac{9}{a^{2}} (a^{2} ((a + 5) (a - 5))) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Rewrite the expression.

$9 ((a + 5) (a - 5)) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Expand $(a + 5) (a - 5)$ using the FOIL Method.

Apply the distributive property.

$9 (a (a - 5) + 5 (a - 5)) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Apply the distributive property.

$9 (a \cdot a + a \cdot - 5 + 5 (a - 5)) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Apply the distributive property.

$9 (a \cdot a + a \cdot - 5 + 5 a + 5 \cdot - 5) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Combine the opposite terms in $a \cdot a + a \cdot - 5 + 5 a + 5 \cdot - 5$ .

Reorder the factors in the terms $a \cdot - 5$ and $5 a$ .

$9 (a \cdot a - 5 a + 5 a + 5 \cdot - 5) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Add $- 5 a$ and $5 a$ .

$9 (a \cdot a + 0 + 5 \cdot - 5) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Add $a \cdot a$ and $0$ .

$9 (a \cdot a + 5 \cdot - 5) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Simplify each term.

Multiply $a$ by $a$ .

$9 (a^{2} + 5 \cdot - 5) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Multiply $5$ by $- 5$ .

$9 (a^{2} - 25) + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Apply the distributive property.

$9 a^{2} + 9 \cdot - 25 + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Multiply $9$ by $- 25$ .

$9 a^{2} - 225 + \frac{16}{(a + 5) (a - 5)} (a^{2} (a + 5) (a - 5)) = 1 (a^{2} (a + 5) (a - 5))$

Cancel the common factor of $(a + 5) (a - 5)$ .

Factor $(a + 5) (a - 5)$ out of $a^{2} (a + 5) (a - 5)$ .

$9 a^{2} - 225 + \frac{16}{(a + 5) (a - 5)} ((a + 5) (a - 5) (a^{2})) = 1 (a^{2} (a + 5) (a - 5))$

Cancel the common factor.

$9 a^{2} - 225 + \frac{16}{(a + 5) (a - 5)} ((a + 5) (a - 5) a^{2}) = 1 (a^{2} (a + 5) (a - 5))$

Rewrite the expression.

$9 a^{2} - 225 + 16 a^{2} = 1 (a^{2} (a + 5) (a - 5))$

Add $9 a^{2}$ and $16 a^{2}$ .

$25 a^{2} - 225 = 1 (a^{2} (a + 5) (a - 5))$

Simplify the right side.

Multiply $a^{2} (a + 5) (a - 5)$ by $1$ .

$25 a^{2} - 225 = a^{2} (a + 5) (a - 5)$

Apply the distributive property.

$25 a^{2} - 225 = (a^{2} a + a^{2} \cdot 5) (a - 5)$

Multiply $a^{2}$ by $a$ by adding the exponents.

Multiply $a^{2}$ by $a$ .

Raise $a$ to the power of $1$ .

$25 a^{2} - 225 = (a^{2} a^{1} + a^{2} \cdot 5) (a - 5)$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$25 a^{2} - 225 = (a^{2 + 1} + a^{2} \cdot 5) (a - 5)$

Add $2$ and $1$ .

$25 a^{2} - 225 = (a^{3} + a^{2} \cdot 5) (a - 5)$

Move $5$ to the left of $a^{2}$ .

$25 a^{2} - 225 = (a^{3} + 5 \cdot a^{2}) (a - 5)$

Expand $(a^{3} + 5 a^{2}) (a - 5)$ using the FOIL Method.

Apply the distributive property.

$25 a^{2} - 225 = a^{3} (a - 5) + 5 a^{2} (a - 5)$

Apply the distributive property.

$25 a^{2} - 225 = a^{3} a + a^{3} \cdot - 5 + 5 a^{2} (a - 5)$

Apply the distributive property.

$25 a^{2} - 225 = a^{3} a + a^{3} \cdot - 5 + 5 a^{2} a + 5 a^{2} \cdot - 5$

Simplify and combine like terms.

Simplify each term.

Multiply $a^{3}$ by $a$ by adding the exponents.

Multiply $a^{3}$ by $a$ .

Raise $a$ to the power of $1$ .

$25 a^{2} - 225 = a^{3} a^{1} + a^{3} \cdot - 5 + 5 a^{2} a + 5 a^{2} \cdot - 5$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$25 a^{2} - 225 = a^{3 + 1} + a^{3} \cdot - 5 + 5 a^{2} a + 5 a^{2} \cdot - 5$

Add $3$ and $1$ .

$25 a^{2} - 225 = a^{4} + a^{3} \cdot - 5 + 5 a^{2} a + 5 a^{2} \cdot - 5$

Move $- 5$ to the left of $a^{3}$ .

$25 a^{2} - 225 = a^{4} - 5 \cdot a^{3} + 5 a^{2} a + 5 a^{2} \cdot - 5$

Multiply $a^{2}$ by $a$ by adding the exponents.

Move $a$ .

$25 a^{2} - 225 = a^{4} - 5 a^{3} + 5 (a \cdot a^{2}) + 5 a^{2} \cdot - 5$

Multiply $a$ by $a^{2}$ .

Raise $a$ to the power of $1$ .

$25 a^{2} - 225 = a^{4} - 5 a^{3} + 5 (a^{1} a^{2}) + 5 a^{2} \cdot - 5$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$25 a^{2} - 225 = a^{4} - 5 a^{3} + 5 a^{1 + 2} + 5 a^{2} \cdot - 5$

Add $1$ and $2$ .

$25 a^{2} - 225 = a^{4} - 5 a^{3} + 5 a^{3} + 5 a^{2} \cdot - 5$

Multiply $- 5$ by $5$ .

$25 a^{2} - 225 = a^{4} - 5 a^{3} + 5 a^{3} - 25 a^{2}$

Add $- 5 a^{3}$ and $5 a^{3}$ .

$25 a^{2} - 225 = a^{4} + 0 - 25 a^{2}$

Add $a^{4}$ and $0$ .

$25 a^{2} - 225 = a^{4} - 25 a^{2}$

Step 4

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.

$a^{4} - 25 a^{2} = 25 a^{2} - 225$

Move all the expressions to the left side of the equation.

Subtract $25 a^{2}$ from both sides of the equation.

$a^{4} - 25 a^{2} - 25 a^{2} = - 225$

Add $225$ to both sides of the equation.

$a^{4} - 25 a^{2} - 25 a^{2} + 225 = 0$

Subtract $25 a^{2}$ from $- 25 a^{2}$ .

$a^{4} - 50 a^{2} + 225 = 0$

Substitute $u = a^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 50 u + 225 = 0$

$u = a^{2}$

Factor $u^{2} - 50 u + 225$ using the AC method.

Consider the form $x^{2} + b x + c$ . 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.

$(u - 45) (u - 5) = 0$

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 $u - 45$ equal to $0$ and solve for $u$ .

Set $u - 45$ equal to $0$ .

$u - 45 = 0$

Add $45$ to both sides of the equation.

$u = 45$

Set $u - 5$ equal to $0$ and solve for $u$ .

Set $u - 5$ equal to $0$ .

$u - 5 = 0$

Add $5$ to both sides of the equation.

$u = 5$

The final solution is all the values that make $(u - 45) (u - 5) = 0$ true.

$u = 45, 5$

Substitute the real value of $u = a^{2}$ back into the solved equation.

$a^{2} = 45$

${(a^{2})}^{1} = 5$

Solve the first equation for $a$ .

$a^{2} = 45$

Solve the equation for $a$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$a = \pm \sqrt{45}$

Simplify $\pm \sqrt{45}$ .

Rewrite $45$ as $3^{2} \cdot 5$ .

Factor $9$ out of $45$ .

$a = \pm \sqrt{9 (5)}$

Rewrite $9$ as $3^{2}$ .

$a = \pm \sqrt{3^{2} \cdot 5}$

Pull terms out from under the radical.

$a = \pm 3 \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$a = 3 \sqrt{5}$

Next, use the negative value of the $\pm$ to find the second solution.

$a = - 3 \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

$a = 3 \sqrt{5}, - 3 \sqrt{5}$

Solve the second equation for $a$ .

${(a^{2})}^{1} = 5$

Solve the equation for $a$ .

Remove parentheses.

$a^{2} = 5$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$a = \pm \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$a = \sqrt{5}$

Next, use the negative value of the $\pm$ to find the second solution.

$a = - \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

$a = \sqrt{5}, - \sqrt{5}$

The solution to $a^{4} - 50 a^{2} + 225 = 0$ is $a = 3 \sqrt{5}, - 3 \sqrt{5}, \sqrt{5}, - \sqrt{5}$ .

$a = 3 \sqrt{5}, - 3 \sqrt{5}, \sqrt{5}, - \sqrt{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$a = 3 \sqrt{5}, - 3 \sqrt{5}, \sqrt{5}, - \sqrt{5}$

Decimal Form:

$a = 6.70820393 \dots, - 6.70820393 \dots, 2.23606797 \dots, - 2.23606797 \dots$

Solve for a 9/(a^2)+16/(a^2-25)=1

Related Questions