Solve for x square root of 8-x=2x+d

$\sqrt{8 - x} = 2 x + d$

Step 1

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

${\sqrt{8 - x}}^{2} = {(2 x + d)}^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8 - x}$ as ${(8 - x)}^{\frac{1}{2}}$ .

${({(8 - x)}^{\frac{1}{2}})}^{2} = {(2 x + d)}^{2}$

Simplify the left side.

Simplify ${({(8 - x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(8 - x)}^{\frac{1}{2}})}^{2}$ .

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

${(8 - x)}^{\frac{1}{2} \cdot 2} = {(2 x + d)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(8 - x)}^{\frac{1}{2} \cdot 2} = {(2 x + d)}^{2}$

Rewrite the expression.

${(8 - x)}^{1} = {(2 x + d)}^{2}$

Simplify.

$8 - x = {(2 x + d)}^{2}$

Simplify the right side.

Simplify ${(2 x + d)}^{2}$ .

Rewrite ${(2 x + d)}^{2}$ as $(2 x + d) (2 x + d)$ .

$8 - x = (2 x + d) (2 x + d)$

Expand $(2 x + d) (2 x + d)$ using the FOIL Method.

Apply the distributive property.

$8 - x = 2 x (2 x + d) + d (2 x + d)$

Apply the distributive property.

$8 - x = 2 x (2 x) + 2 x d + d (2 x + d)$

Apply the distributive property.

$8 - x = 2 x (2 x) + 2 x d + d (2 x) + d \cdot d$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$8 - x = 2 \cdot 2 x \cdot x + 2 x d + d (2 x) + d \cdot d$

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

Move $x$ .

$8 - x = 2 \cdot 2 (x \cdot x) + 2 x d + d (2 x) + d \cdot d$

Multiply $x$ by $x$ .

$8 - x = 2 \cdot 2 x^{2} + 2 x d + d (2 x) + d \cdot d$

Multiply $2$ by $2$ .

$8 - x = 4 x^{2} + 2 x d + d (2 x) + d \cdot d$

Rewrite using the commutative property of multiplication.

$8 - x = 4 x^{2} + 2 x d + 2 d x + d \cdot d$

Multiply $d$ by $d$ .

$8 - x = 4 x^{2} + 2 x d + 2 d x + d^{2}$

Add $2 x d$ and $2 d x$ .

Move $x$ .

$8 - x = 4 x^{2} + 2 d x + 2 d x + d^{2}$

Add $2 d x$ and $2 d x$ .

$8 - x = 4 x^{2} + 4 d x + d^{2}$

Step 3

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$4 x^{2} + 4 d x + d^{2} = 8 - x$

Add $x$ to both sides of the equation.

$4 x^{2} + 4 d x + d^{2} + x = 8$

Subtract $8$ from both sides of the equation.

$4 x^{2} + 4 d x + d^{2} + x - 8 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 4$ , $b = 4 d + 1$ , and $c = d^{2} - 8$ into the quadratic formula and solve for $x$ .

$\frac{- (4 d + 1) \pm \sqrt{{(4 d + 1)}^{2} - 4 \cdot (4 \cdot (d^{2} - 8))}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Apply the distributive property.

$x = \frac{- (4 d) - 1 \cdot 1 \pm \sqrt{{(4 d + 1)}^{2} - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $4$ by $- 1$ .

$x = \frac{- 4 d - 1 \cdot 1 \pm \sqrt{{(4 d + 1)}^{2} - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $- 1$ by $1$ .

$x = \frac{- 4 d - 1 \pm \sqrt{{(4 d + 1)}^{2} - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Rewrite ${(4 d + 1)}^{2}$ as $(4 d + 1) (4 d + 1)$ .

$x = \frac{- 4 d - 1 \pm \sqrt{(4 d + 1) (4 d + 1) - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Expand $(4 d + 1) (4 d + 1)$ using the FOIL Method.

Apply the distributive property.

$x = \frac{- 4 d - 1 \pm \sqrt{4 d (4 d + 1) + 1 (4 d + 1) - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Apply the distributive property.

$x = \frac{- 4 d - 1 \pm \sqrt{4 d (4 d) + 4 d \cdot 1 + 1 (4 d + 1) - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Apply the distributive property.

$x = \frac{- 4 d - 1 \pm \sqrt{4 d (4 d) + 4 d \cdot 1 + 1 (4 d) + 1 \cdot 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$x = \frac{- 4 d - 1 \pm \sqrt{4 \cdot (4 d \cdot d) + 4 d \cdot 1 + 1 (4 d) + 1 \cdot 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $d$ by $d$ by adding the exponents.

Move $d$ .

$x = \frac{- 4 d - 1 \pm \sqrt{4 \cdot (4 (d \cdot d)) + 4 d \cdot 1 + 1 (4 d) + 1 \cdot 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $d$ by $d$ .

$x = \frac{- 4 d - 1 \pm \sqrt{4 \cdot (4 d^{2}) + 4 d \cdot 1 + 1 (4 d) + 1 \cdot 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $4$ by $4$ .

$x = \frac{- 4 d - 1 \pm \sqrt{16 d^{2} + 4 d \cdot 1 + 1 (4 d) + 1 \cdot 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $4$ by $1$ .

$x = \frac{- 4 d - 1 \pm \sqrt{16 d^{2} + 4 d + 1 (4 d) + 1 \cdot 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $4 d$ by $1$ .

$x = \frac{- 4 d - 1 \pm \sqrt{16 d^{2} + 4 d + 4 d + 1 \cdot 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $1$ by $1$ .

$x = \frac{- 4 d - 1 \pm \sqrt{16 d^{2} + 4 d + 4 d + 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Add $4 d$ and $4 d$ .

$x = \frac{- 4 d - 1 \pm \sqrt{16 d^{2} + 8 d + 1 - 4 \cdot 4 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$x = \frac{- 4 d - 1 \pm \sqrt{16 d^{2} + 8 d + 1 - 16 \cdot (d^{2} - 8)}}{2 \cdot 4}$

Apply the distributive property.

$x = \frac{- 4 d - 1 \pm \sqrt{16 d^{2} + 8 d + 1 - 16 d^{2} - 16 \cdot - 8}}{2 \cdot 4}$

Multiply $- 16$ by $- 8$ .

$x = \frac{- 4 d - 1 \pm \sqrt{16 d^{2} + 8 d + 1 - 16 d^{2} + 128}}{2 \cdot 4}$

Subtract $16 d^{2}$ from $16 d^{2}$ .

$x = \frac{- 4 d - 1 \pm \sqrt{8 d + 1 + 0 + 128}}{2 \cdot 4}$

Add $8 d$ and $0$ .

$x = \frac{- 4 d - 1 \pm \sqrt{8 d + 1 + 128}}{2 \cdot 4}$

Add $1$ and $128$ .

$x = \frac{- 4 d - 1 \pm \sqrt{8 d + 129}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 4 d - 1 \pm \sqrt{8 d + 129}}{8}$

The final answer is the combination of both solutions.

$x = - \frac{4 d + 1 - \sqrt{8 d + 129}}{8}$

$x = - \frac{4 d + 1 + \sqrt{8 d + 129}}{8}$

$x = - \frac{4 d + 1 - \sqrt{8 d + 129}}{8}$

$x = - \frac{4 d + 1 + \sqrt{8 d + 129}}{8}$

Solve for x square root of 8-x=2x+d

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