Solve sqrt{6+2k}=1-k

$$\sqrt{ 6+2k } = 1-k$$

Solve for k

$$k=-1$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{6+2k} to the power of 2 and get 6+2k.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-k\right)^{2}.

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

Subtract 1 from both sides.

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

Subtract 1 from 6 to get 5.

$$5+2k=-2k+k^{2}$$

Add 2k to both sides.

$$5+2k+2k=k^{2}$$

Combine 2k and 2k to get 4k.

$$5+4k=k^{2}$$

Subtract k^{2} from both sides.

$$5+4k-k^{2}=0$$

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

$$-k^{2}+4k+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -k^{2}+ak+bk+5. To find a and b, set up a system to be solved.

$$a+b=4 ab=-5=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=5 b=-1$$

Rewrite -k^{2}+4k+5 as \left(-k^{2}+5k\right)+\left(-k+5\right).

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

Factor out -k in the first and -1 in the second group.

$$-k\left(k-5\right)-\left(k-5\right)$$

Factor out common term k-5 by using distributive property.

$$\left(k-5\right)\left(-k-1\right)$$

To find equation solutions, solve k-5=0 and -k-1=0.

$$k=5 k=-1$$

Substitute 5 for k in the equation \sqrt{6+2k}=1-k.

$$\sqrt{6+2\times 5}=1-5$$

Simplify. The value k=5 does not satisfy the equation because the left and the right hand side have opposite signs.

$$4=-4$$

Substitute -1 for k in the equation \sqrt{6+2k}=1-k.

$$\sqrt{6+2\left(-1\right)}=1-\left(-1\right)$$

Simplify. The value k=-1 satisfies the equation.

$$2=2$$

Equation \sqrt{2k+6}=1-k has a unique solution.

$$k=-1$$