Solve sqrt{x+6}-sqrt{5x+1}+1=0

$$\sqrt { x + 6 } - \sqrt { 5 x + 1 } + 1 = 0$$

Solve for x

$$x=3$$

Solution Steps

Subtract -\sqrt{5x+1}+1 from both sides of the equation.

$$\sqrt{x+6}=-\left(-\sqrt{5x+1}+1\right)$$

To find the opposite of -\sqrt{5x+1}+1, find the opposite of each term.

$$\sqrt{x+6}=-\left(-\sqrt{5x+1}\right)-1$$

The opposite of -\sqrt{5x+1} is \sqrt{5x+1}.

$$\sqrt{x+6}=\sqrt{5x+1}-1$$

Square both sides of the equation.

$$\left(\sqrt{x+6}\right)^{2}=\left(\sqrt{5x+1}-1\right)^{2}$$

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

$$x+6=\left(\sqrt{5x+1}-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5x+1}-1\right)^{2}.

$$x+6=\left(\sqrt{5x+1}\right)^{2}-2\sqrt{5x+1}+1$$

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

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

Add 1 and 1 to get 2.

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

Subtract 5x+2 from both sides of the equation.

$$x+6-\left(5x+2\right)=-2\sqrt{5x+1}$$

To find the opposite of 5x+2, find the opposite of each term.

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

Combine x and -5x to get -4x.

$$-4x+6-2=-2\sqrt{5x+1}$$

Subtract 2 from 6 to get 4.

$$-4x+4=-2\sqrt{5x+1}$$

Square both sides of the equation.

$$\left(-4x+4\right)^{2}=\left(-2\sqrt{5x+1}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-4x+4\right)^{2}.

$$16x^{2}-32x+16=\left(-2\sqrt{5x+1}\right)^{2}$$

Expand \left(-2\sqrt{5x+1}\right)^{2}.

$$16x^{2}-32x+16=\left(-2\right)^{2}\left(\sqrt{5x+1}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$16x^{2}-32x+16=4\left(\sqrt{5x+1}\right)^{2}$$

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

$$16x^{2}-32x+16=4\left(5x+1\right)$$

Use the distributive property to multiply 4 by 5x+1.

$$16x^{2}-32x+16=20x+4$$

Subtract 20x from both sides.

$$16x^{2}-32x+16-20x=4$$

Combine -32x and -20x to get -52x.

$$16x^{2}-52x+16=4$$

Subtract 4 from both sides.

$$16x^{2}-52x+16-4=0$$

Subtract 4 from 16 to get 12.

$$16x^{2}-52x+12=0$$

Divide both sides by 4.

$$4x^{2}-13x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-13 ab=4\times 3=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -13.

$$a=-12 b=-1$$

Rewrite 4x^{2}-13x+3 as \left(4x^{2}-12x\right)+\left(-x+3\right).

$$\left(4x^{2}-12x\right)+\left(-x+3\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(4x-1\right)$$

To find equation solutions, solve x-3=0 and 4x-1=0.

$$x=3 x=\frac{1}{4}$$

Substitute 3 for x in the equation \sqrt{x+6}-\sqrt{5x+1}+1=0.

$$\sqrt{3+6}-\sqrt{5\times 3+1}+1=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Substitute \frac{1}{4} for x in the equation \sqrt{x+6}-\sqrt{5x+1}+1=0.

$$\sqrt{\frac{1}{4}+6}-\sqrt{5\times \frac{1}{4}+1}+1=0$$

Simplify. The value x=\frac{1}{4} does not satisfy the equation.

$$2=0$$

Substitute 3 for x in the equation \sqrt{x+6}-\sqrt{5x+1}+1=0.

$$\sqrt{3+6}-\sqrt{5\times 3+1}+1=0$$

Simplify. The value x=3 satisfies the equation.

$$0=0$$

Equation \sqrt{x+6}=\sqrt{5x+1}-1 has a unique solution.

$$x=3$$