Solve 3-k=sqrt{3k+1}

$$3 - k = \sqrt { 3 k + 1 }$$

Solve for k

$$k=1$$

Solution Steps

Square both sides of the equation.

$$\left(3-k\right)^{2}=\left(\sqrt{3k+1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-k\right)^{2}.

$$9-6k+k^{2}=\left(\sqrt{3k+1}\right)^{2}$$

Calculate \sqrt{3k+1} to the power of 2 and get 3k+1.

$$9-6k+k^{2}=3k+1$$

Subtract 3k from both sides.

$$9-6k+k^{2}-3k=1$$

Combine -6k and -3k to get -9k.

$$9-9k+k^{2}=1$$

Subtract 1 from both sides.

$$9-9k+k^{2}-1=0$$

Subtract 1 from 9 to get 8.

$$8-9k+k^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$k^{2}-9k+8=0$$

To solve the equation, factor k^{2}-9k+8 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.

$$a+b=-9 ab=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -9.

$$a=-8 b=-1$$

Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.

$$\left(k-8\right)\left(k-1\right)$$

To find equation solutions, solve k-8=0 and k-1=0.

$$k=8 k=1$$

Substitute 8 for k in the equation 3-k=\sqrt{3k+1}.

$$3-8=\sqrt{3\times 8+1}$$

Simplify. The value k=8 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-5=5$$

Substitute 1 for k in the equation 3-k=\sqrt{3k+1}.

$$3-1=\sqrt{3\times 1+1}$$

Simplify. The value k=1 satisfies the equation.

$$2=2$$

Equation 3-k=\sqrt{3k+1} has a unique solution.

$$k=1$$