Find the Roots (Zeros) f(x)=4 square root of 2x-1+3

$f (x) = 4 \sqrt{2 x - 1} + 3$

Set $4 \sqrt{2 x - 1} + 3$ equal to $0$ .

$4 \sqrt{2 x - 1} + 3 = 0$

Solve for $x$ .

Subtract $3$ from both sides of the equation.

$4 \sqrt{2 x - 1} = - 3$

To remove the radical on the left side of the equation, square both sides of the equation.

${(4 \sqrt{2 x - 1})}^{2} = {(- 3)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 1}$ as ${(2 x - 1)}^{\frac{1}{2}}$ .

${(4 {(2 x - 1)}^{\frac{1}{2}})}^{2} = {(- 3)}^{2}$

Simplify the left side.

Simplify ${(4 {(2 x - 1)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $4 {(2 x - 1)}^{\frac{1}{2}}$ .

$4^{2} {({(2 x - 1)}^{\frac{1}{2}})}^{2} = {(- 3)}^{2}$

Raise $4$ to the power of $2$ .

$16 {({(2 x - 1)}^{\frac{1}{2}})}^{2} = {(- 3)}^{2}$

Multiply the exponents in ${({(2 x - 1)}^{\frac{1}{2}})}^{2}$ .

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

$16 {(2 x - 1)}^{\frac{1}{2} \cdot 2} = {(- 3)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$16 {(2 x - 1)}^{\frac{1}{2} \cdot 2} = {(- 3)}^{2}$

Rewrite the expression.

$16 {(2 x - 1)}^{1} = {(- 3)}^{2}$

Simplify.

$16 (2 x - 1) = {(- 3)}^{2}$

Apply the distributive property.

$16 (2 x) + 16 \cdot - 1 = {(- 3)}^{2}$

Multiply.

Multiply $2$ by $16$ .

$32 x + 16 \cdot - 1 = {(- 3)}^{2}$

Multiply $16$ by $- 1$ .

$32 x - 16 = {(- 3)}^{2}$

Simplify the right side.

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

$32 x - 16 = 9$

Solve for $x$ .

Move all terms not containing $x$ to the right side of the equation.

Add $16$ to both sides of the equation.

$32 x = 9 + 16$

Add $9$ and $16$ .

$32 x = 25$

Divide each term in $32 x = 25$ by $32$ and simplify.

Divide each term in $32 x = 25$ by $32$ .

$\frac{32 x}{32} = \frac{25}{32}$

Simplify the left side.

Cancel the common factor of $32$ .

Cancel the common factor.

$\frac{32 x}{32} = \frac{25}{32}$

Divide $x$ by $1$ .

$x = \frac{25}{32}$

Exclude the solutions that do not make $4 \sqrt{2 x - 1} + 3 = 0$ true.

$No solution$

Find the Roots (Zeros) f(x)=4 square root of 2x-1+3

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