Find the Roots (Zeros) F(x)=(4x-17) natural log of (2x-9)(x-4)

$F (x) = (4 x - 17) \ln ((2 x - 9) (x - 4))$

Set $(4 x - 17) \ln ((2 x - 9) (x - 4))$ equal to $0$ .

$(4 x - 17) \ln ((2 x - 9) (x - 4)) = 0$

Solve for $x$ .

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

$4 x - 17 = 0$

$\ln ((2 x - 9) (x - 4)) = 0$

Set $4 x - 17$ equal to $0$ and solve for $x$ .

Set $4 x - 17$ equal to $0$ .

$4 x - 17 = 0$

Solve $4 x - 17 = 0$ for $x$ .

Add $17$ to both sides of the equation.

$4 x = 17$

Divide each term in $4 x = 17$ by $4$ and simplify.

Divide each term in $4 x = 17$ by $4$ .

$\frac{4 x}{4} = \frac{17}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{17}{4}$

Divide $x$ by $1$ .

$x = \frac{17}{4}$

Set $\ln ((2 x - 9) (x - 4))$ equal to $0$ and solve for $x$ .

Set $\ln ((2 x - 9) (x - 4))$ equal to $0$ .

$\ln ((2 x - 9) (x - 4)) = 0$

Solve $\ln ((2 x - 9) (x - 4)) = 0$ for $x$ .

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln ((2 x - 9) (x - 4))} = e^{0}$

Rewrite $\ln ((2 x - 9) (x - 4)) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = (2 x - 9) (x - 4)$

Solve for $x$ .

Rewrite the equation as $(2 x - 9) (x - 4) = e^{0}$ .

$(2 x - 9) (x - 4) = e^{0}$

Simplify $(2 x - 9) (x - 4)$ .

Expand $(2 x - 9) (x - 4)$ using the FOIL Method.

Apply the distributive property.

$2 x (x - 4) - 9 (x - 4) = e^{0}$

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 4 - 9 (x - 4) = e^{0}$

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 4 - 9 x - 9 \cdot - 4 = e^{0}$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 4 - 9 x - 9 \cdot - 4 = e^{0}$

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 4 - 9 x - 9 \cdot - 4 = e^{0}$

Multiply $- 4$ by $2$ .

$2 x^{2} - 8 x - 9 x - 9 \cdot - 4 = e^{0}$

Multiply $- 9$ by $- 4$ .

$2 x^{2} - 8 x - 9 x + 36 = e^{0}$

Subtract $9 x$ from $- 8 x$ .

$2 x^{2} - 17 x + 36 = e^{0}$

Anything raised to $0$ is $1$ .

$2 x^{2} - 17 x + 36 = 1$

Subtract $1$ from both sides of the equation.

$2 x^{2} - 17 x + 36 - 1 = 0$

Subtract $1$ from $36$ .

$2 x^{2} - 17 x + 35 = 0$

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 $a \cdot c = 2 \cdot 35 = 70$ and whose sum is $b = - 17$ .

Factor $- 17$ out of $- 17 x$ .

$2 x^{2} - 17 x + 35 = 0$

Rewrite $- 17$ as $- 7$ plus $- 10$

$2 x^{2} + (- 7 - 10) x + 35 = 0$

Apply the distributive property.

$2 x^{2} - 7 x - 10 x + 35 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 7 x) - 10 x + 35 = 0$

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

$x (2 x - 7) - 5 (2 x - 7) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 7$ .

$(2 x - 7) (x - 5) = 0$

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

$2 x - 7 = 0$

$x - 5 = 0$

Set $2 x - 7$ equal to $0$ and solve for $x$ .

Set $2 x - 7$ equal to $0$ .

$2 x - 7 = 0$

Solve $2 x - 7 = 0$ for $x$ .

Add $7$ to both sides of the equation.

$2 x = 7$

Divide each term in $2 x = 7$ by $2$ and simplify.

Divide each term in $2 x = 7$ by $2$ .

$\frac{2 x}{2} = \frac{7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Divide $x$ by $1$ .

$x = \frac{7}{2}$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $(2 x - 7) (x - 5) = 0$ true.

$x = \frac{7}{2}, 5$

The final solution is all the values that make $(4 x - 17) \ln ((2 x - 9) (x - 4)) = 0$ true.

$x = \frac{17}{4}, \frac{7}{2}, 5$

Exclude the solutions that do not make $(4 x - 17) \ln ((2 x - 9) (x - 4)) = 0$ true.

$x = \frac{7}{2}, 5$

Find the Roots (Zeros) F(x)=(4x-17) natural log of (2x-9)(x-4)

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