Solve by Factoring 4k^(4/3)-37k^(2/3)+9=0

$4 k^{\frac{4}{3}} - 37 k^{\frac{2}{3}} + 9 = 0$

Step 1

Rewrite $k^{\frac{4}{3}}$ as ${(k^{\frac{2}{3}})}^{2}$ .

$4 {(k^{\frac{2}{3}})}^{2} - 37 k^{\frac{2}{3}} + 9 = 0$

Step 2

Let $u = k^{\frac{2}{3}}$ . Substitute $u$ for all occurrences of $k^{\frac{2}{3}}$ .

$4 u^{2} - 37 u + 9 = 0$

Step 3

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 9 = 36$ and whose sum is $b = - 37$ .

Factor $- 37$ out of $- 37 u$ .

$4 u^{2} - 37 u + 9 = 0$

Rewrite $- 37$ as $- 1$ plus $- 36$

$4 u^{2} + (- 1 - 36) u + 9 = 0$

Apply the distributive property.

$4 u^{2} - 1 u - 36 u + 9 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 u^{2} - 1 u) - 36 u + 9 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (4 u - 1) - 9 (4 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$(4 u - 1) (u - 9) = 0$

Step 4

Replace all occurrences of $u$ with $k^{\frac{2}{3}}$ .

$(4 k^{\frac{2}{3}} - 1) (k^{\frac{2}{3}} - 9) = 0$

Step 5

Rewrite $4 k^{\frac{2}{3}}$ as ${(2 k^{\frac{1}{3}})}^{2}$ .

$({(2 k^{\frac{1}{3}})}^{2} - 1) (k^{\frac{2}{3}} - 9) = 0$

Step 6

Rewrite $1$ as $1^{2}$ .

$({(2 k^{\frac{1}{3}})}^{2} - 1^{2}) (k^{\frac{2}{3}} - 9) = 0$

Step 7

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 k^{\frac{1}{3}}$ and $b = 1$ .

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) (k^{\frac{2}{3}} - 9) = 0$

Step 8

Rewrite $k^{\frac{2}{3}}$ as ${(k^{\frac{1}{3}})}^{2}$ .

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) ({(k^{\frac{1}{3}})}^{2} - 9) = 0$

Step 9

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

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) ({(k^{\frac{1}{3}})}^{2} - 3^{2}) = 0$

Step 10

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = k^{\frac{1}{3}}$ and $b = 3$ .

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) ((k^{\frac{1}{3}} + 3) (k^{\frac{1}{3}} - 3)) = 0$

Remove unnecessary parentheses.

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) (k^{\frac{1}{3}} + 3) (k^{\frac{1}{3}} - 3) = 0$

Step 11

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

$2 k^{\frac{1}{3}} + 1 = 0$

$2 k^{\frac{1}{3}} - 1 = 0$

$k^{\frac{1}{3}} + 3 = 0$

$k^{\frac{1}{3}} - 3 = 0$

Step 12

Set $2 k^{\frac{1}{3}} + 1$ equal to $0$ and solve for $k$ .

Set $2 k^{\frac{1}{3}} + 1$ equal to $0$ .

$2 k^{\frac{1}{3}} + 1 = 0$

Solve $2 k^{\frac{1}{3}} + 1 = 0$ for $k$ .

Subtract $1$ from both sides of the equation.

$2 k^{\frac{1}{3}} = - 1$

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

${(2 k^{\frac{1}{3}})}^{3} = {(- 1)}^{3}$

Simplify the exponent.

Simplify the left side.

Simplify ${(2 k^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $2 k^{\frac{1}{3}}$ .

$2^{3} {(k^{\frac{1}{3}})}^{3} = {(- 1)}^{3}$

Raise $2$ to the power of $3$ .

$8 {(k^{\frac{1}{3}})}^{3} = {(- 1)}^{3}$

Multiply the exponents in ${(k^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$8 k^{\frac{1}{3} \cdot 3} = {(- 1)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8 k^{\frac{1}{3} \cdot 3} = {(- 1)}^{3}$

Rewrite the expression.

$8 k^{1} = {(- 1)}^{3}$

Simplify.

$8 k = {(- 1)}^{3}$

Simplify the right side.

Raise $- 1$ to the power of $3$ .

$8 k = - 1$

Divide each term in $8 k = - 1$ by $8$ and simplify.

Divide each term in $8 k = - 1$ by $8$ .

$\frac{8 k}{8} = \frac{- 1}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 k}{8} = \frac{- 1}{8}$

Divide $k$ by $1$ .

$k = \frac{- 1}{8}$

Simplify the right side.

Move the negative in front of the fraction.

$k = - \frac{1}{8}$

Step 13

Set $2 k^{\frac{1}{3}} - 1$ equal to $0$ and solve for $k$ .

Set $2 k^{\frac{1}{3}} - 1$ equal to $0$ .

$2 k^{\frac{1}{3}} - 1 = 0$

Solve $2 k^{\frac{1}{3}} - 1 = 0$ for $k$ .

Add $1$ to both sides of the equation.

$2 k^{\frac{1}{3}} = 1$

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

${(2 k^{\frac{1}{3}})}^{3} = 1^{3}$

Simplify the exponent.

Simplify the left side.

Simplify ${(2 k^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $2 k^{\frac{1}{3}}$ .

$2^{3} {(k^{\frac{1}{3}})}^{3} = 1^{3}$

Raise $2$ to the power of $3$ .

$8 {(k^{\frac{1}{3}})}^{3} = 1^{3}$

Multiply the exponents in ${(k^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$8 k^{\frac{1}{3} \cdot 3} = 1^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8 k^{\frac{1}{3} \cdot 3} = 1^{3}$

Rewrite the expression.

$8 k^{1} = 1^{3}$

Simplify.

$8 k = 1^{3}$

Simplify the right side.

One to any power is one.

$8 k = 1$

Divide each term in $8 k = 1$ by $8$ and simplify.

Divide each term in $8 k = 1$ by $8$ .

$\frac{8 k}{8} = \frac{1}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 k}{8} = \frac{1}{8}$

Divide $k$ by $1$ .

$k = \frac{1}{8}$

Step 14

Set $k^{\frac{1}{3}} + 3$ equal to $0$ and solve for $k$ .

Set $k^{\frac{1}{3}} + 3$ equal to $0$ .

$k^{\frac{1}{3}} + 3 = 0$

Solve $k^{\frac{1}{3}} + 3 = 0$ for $k$ .

Subtract $3$ from both sides of the equation.

$k^{\frac{1}{3}} = - 3$

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

${(k^{\frac{1}{3}})}^{3} = {(- 3)}^{3}$

Simplify the exponent.

Simplify the left side.

Simplify ${(k^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${(k^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$k^{\frac{1}{3} \cdot 3} = {(- 3)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$k^{\frac{1}{3} \cdot 3} = {(- 3)}^{3}$

Rewrite the expression.

$k^{1} = {(- 3)}^{3}$

Simplify.

$k = {(- 3)}^{3}$

Simplify the right side.

Raise $- 3$ to the power of $3$ .

$k = - 27$

Step 15

Set $k^{\frac{1}{3}} - 3$ equal to $0$ and solve for $k$ .

Set $k^{\frac{1}{3}} - 3$ equal to $0$ .

$k^{\frac{1}{3}} - 3 = 0$

Solve $k^{\frac{1}{3}} - 3 = 0$ for $k$ .

Add $3$ to both sides of the equation.

$k^{\frac{1}{3}} = 3$

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

${(k^{\frac{1}{3}})}^{3} = 3^{3}$

Simplify the exponent.

Simplify the left side.

Simplify ${(k^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${(k^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$k^{\frac{1}{3} \cdot 3} = 3^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$k^{\frac{1}{3} \cdot 3} = 3^{3}$

Rewrite the expression.

$k^{1} = 3^{3}$

Simplify.

$k = 3^{3}$

Simplify the right side.

Raise $3$ to the power of $3$ .

$k = 27$

Step 16

The final solution is all the values that make $(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) (k^{\frac{1}{3}} + 3) (k^{\frac{1}{3}} - 3) = 0$ true.

$k = - \frac{1}{8}, \frac{1}{8}, - 27, 27$

Solve by Factoring 4k^(4/3)-37k^(2/3)+9=0

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