Solve sqrt{5k+2}+8=11*k

$$\sqrt { 5 k + 2 } + 8 = 11 \cdot k$$

Solve for k

$$k=\frac{\sqrt{2753}+181}{242}\approx 0.964748093$$

Solution Steps

Subtract 8 from both sides of the equation.

$$\sqrt{5k+2}=11k-8$$

Square both sides of the equation.

$$\left(\sqrt{5k+2}\right)^{2}=\left(11k-8\right)^{2}$$

Calculate \sqrt{5k+2} to the power of 2 and get 5k+2.

$$5k+2=\left(11k-8\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(11k-8\right)^{2}.

$$5k+2=121k^{2}-176k+64$$

Subtract 121k^{2} from both sides.

$$5k+2-121k^{2}=-176k+64$$

Add 176k to both sides.

$$5k+2-121k^{2}+176k=64$$

Combine 5k and 176k to get 181k.

$$181k+2-121k^{2}=64$$

Subtract 64 from both sides.

$$181k+2-121k^{2}-64=0$$

Subtract 64 from 2 to get -62.

$$181k-62-121k^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-121k^{2}+181k-62=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -121 for a, 181 for b, and -62 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-181±\sqrt{181^{2}-4\left(-121\right)\left(-62\right)}}{2\left(-121\right)}$$

Square 181.

$$k=\frac{-181±\sqrt{32761-4\left(-121\right)\left(-62\right)}}{2\left(-121\right)}$$

Multiply -4 times -121.

$$k=\frac{-181±\sqrt{32761+484\left(-62\right)}}{2\left(-121\right)}$$

Multiply 484 times -62.

$$k=\frac{-181±\sqrt{32761-30008}}{2\left(-121\right)}$$

Add 32761 to -30008.

$$k=\frac{-181±\sqrt{2753}}{2\left(-121\right)}$$

Multiply 2 times -121.

$$k=\frac{-181±\sqrt{2753}}{-242}$$

Now solve the equation k=\frac{-181±\sqrt{2753}}{-242} when ± is plus. Add -181 to \sqrt{2753}.

$$k=\frac{\sqrt{2753}-181}{-242}$$

Divide -181+\sqrt{2753} by -242.

$$k=\frac{181-\sqrt{2753}}{242}$$

Now solve the equation k=\frac{-181±\sqrt{2753}}{-242} when ± is minus. Subtract \sqrt{2753} from -181.

$$k=\frac{-\sqrt{2753}-181}{-242}$$

Divide -181-\sqrt{2753} by -242.

$$k=\frac{\sqrt{2753}+181}{242}$$

The equation is now solved.

$$k=\frac{181-\sqrt{2753}}{242} k=\frac{\sqrt{2753}+181}{242}$$

Substitute \frac{181-\sqrt{2753}}{242} for k in the equation \sqrt{5k+2}+8=11k.

$$\sqrt{5\times \frac{181-\sqrt{2753}}{242}+2}+8=11\times \frac{181-\sqrt{2753}}{242}$$

Simplify. The value k=\frac{181-\sqrt{2753}}{242} does not satisfy the equation.

$$\frac{171}{22}+\frac{1}{22}\times 2753^{\frac{1}{2}}=\frac{181}{22}-\frac{1}{22}\times 2753^{\frac{1}{2}}$$

Substitute \frac{\sqrt{2753}+181}{242} for k in the equation \sqrt{5k+2}+8=11k.

$$\sqrt{5\times \frac{\sqrt{2753}+181}{242}+2}+8=11\times \frac{\sqrt{2753}+181}{242}$$

Simplify. The value k=\frac{\sqrt{2753}+181}{242} satisfies the equation.

$$\frac{181}{22}+\frac{1}{22}\times 2753^{\frac{1}{2}}=\frac{1}{22}\times 2753^{\frac{1}{2}}+\frac{181}{22}$$

Equation \sqrt{5k+2}=11k-8 has a unique solution.

$$k=\frac{\sqrt{2753}+181}{242}$$