Solve for z 6- square root of z^2-z=8-z

$6 - \sqrt{z^{2} - z} = 8 - z$

Step 1

Move all terms not containing $- \sqrt{z^{2} - z}$ to the right side of the equation.

Subtract $6$ from both sides of the equation.

$- \sqrt{z^{2} - z} = 8 - z - 6$

Subtract $6$ from $8$ .

$- \sqrt{z^{2} - z} = - z + 2$

Step 2

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

${(- \sqrt{z^{2} - z})}^{2} = {(- z + 2)}^{2}$

Step 3

Simplify each side of the equation.

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

${(- {(z^{2} - z)}^{\frac{1}{2}})}^{2} = {(- z + 2)}^{2}$

Simplify the left side.

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

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

${(- 1)}^{2} {({(z^{2} - z)}^{\frac{1}{2}})}^{2} = {(- z + 2)}^{2}$

Raise $- 1$ to the power of $2$ .

$1 {({(z^{2} - z)}^{\frac{1}{2}})}^{2} = {(- z + 2)}^{2}$

Multiply ${({(z^{2} - z)}^{\frac{1}{2}})}^{2}$ by $1$ .

${({(z^{2} - z)}^{\frac{1}{2}})}^{2} = {(- z + 2)}^{2}$

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

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

${(z^{2} - z)}^{\frac{1}{2} \cdot 2} = {(- z + 2)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(z^{2} - z)}^{\frac{1}{2} \cdot 2} = {(- z + 2)}^{2}$

Rewrite the expression.

${(z^{2} - z)}^{1} = {(- z + 2)}^{2}$

Simplify.

$z^{2} - z = {(- z + 2)}^{2}$

Simplify the right side.

Simplify ${(- z + 2)}^{2}$ .

Rewrite ${(- z + 2)}^{2}$ as $(- z + 2) (- z + 2)$ .

$z^{2} - z = (- z + 2) (- z + 2)$

Expand $(- z + 2) (- z + 2)$ using the FOIL Method.

Apply the distributive property.

$z^{2} - z = - z (- z + 2) + 2 (- z + 2)$

Apply the distributive property.

$z^{2} - z = - z (- z) - z \cdot 2 + 2 (- z + 2)$

Apply the distributive property.

$z^{2} - z = - z (- z) - z \cdot 2 + 2 (- z) + 2 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$z^{2} - z = - 1 \cdot - 1 z \cdot z - z \cdot 2 + 2 (- z) + 2 \cdot 2$

Multiply $z$ by $z$ by adding the exponents.

Move $z$ .

$z^{2} - z = - 1 \cdot - 1 (z \cdot z) - z \cdot 2 + 2 (- z) + 2 \cdot 2$

Multiply $z$ by $z$ .

$z^{2} - z = - 1 \cdot - 1 z^{2} - z \cdot 2 + 2 (- z) + 2 \cdot 2$

Multiply $- 1$ by $- 1$ .

$z^{2} - z = 1 z^{2} - z \cdot 2 + 2 (- z) + 2 \cdot 2$

Multiply $z^{2}$ by $1$ .

$z^{2} - z = z^{2} - z \cdot 2 + 2 (- z) + 2 \cdot 2$

Multiply $2$ by $- 1$ .

$z^{2} - z = z^{2} - 2 z + 2 (- z) + 2 \cdot 2$

Multiply $- 1$ by $2$ .

$z^{2} - z = z^{2} - 2 z - 2 z + 2 \cdot 2$

Multiply $2$ by $2$ .

$z^{2} - z = z^{2} - 2 z - 2 z + 4$

Subtract $2 z$ from $- 2 z$ .

$z^{2} - z = z^{2} - 4 z + 4$

Step 4

Solve for $z$ .

Move all terms containing $z$ to the left side of the equation.

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

$z^{2} - z - z^{2} = - 4 z + 4$

Add $4 z$ to both sides of the equation.

$z^{2} - z - z^{2} + 4 z = 4$

Combine the opposite terms in $z^{2} - z - z^{2} + 4 z$ .

Subtract $z^{2}$ from $z^{2}$ .

$- z + 0 + 4 z = 4$

Add $- z$ and $0$ .

$- z + 4 z = 4$

Add $- z$ and $4 z$ .

$3 z = 4$

Divide each term in $3 z = 4$ by $3$ and simplify.

Divide each term in $3 z = 4$ by $3$ .

$\frac{3 z}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 z}{3} = \frac{4}{3}$

Divide $z$ by $1$ .

$z = \frac{4}{3}$

Step 5

Exclude the solutions that do not make $6 - \sqrt{z^{2} - z} = 8 - z$ true.

$No solution$

Solve for z 6- square root of z^2-z=8-z

Related Questions

Support LCMGCF.com by disabling your adblocker.