Solve sqrt{x+5}+1=2sqrt{x}quad

$$\sqrt { x + 5 } + 1 = 2 \sqrt { x } \quad$$

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Add 5 and 1 to get 6.

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

Expand \left(2\sqrt{x}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$x+6+2\sqrt{x+5}=4\left(\sqrt{x}\right)^{2}$$

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

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

Subtract x+6 from both sides of the equation.

$$2\sqrt{x+5}=4x-\left(x+6\right)$$

To find the opposite of x+6, find the opposite of each term.

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

Combine 4x and -x to get 3x.

$$2\sqrt{x+5}=3x-6$$

Square both sides of the equation.

$$\left(2\sqrt{x+5}\right)^{2}=\left(3x-6\right)^{2}$$

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

$$2^{2}\left(\sqrt{x+5}\right)^{2}=\left(3x-6\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x+5}\right)^{2}=\left(3x-6\right)^{2}$$

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

$$4\left(x+5\right)=\left(3x-6\right)^{2}$$

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

$$4x+20=\left(3x-6\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-6\right)^{2}.

$$4x+20=9x^{2}-36x+36$$

Subtract 9x^{2} from both sides.

$$4x+20-9x^{2}=-36x+36$$

Add 36x to both sides.

$$4x+20-9x^{2}+36x=36$$

Combine 4x and 36x to get 40x.

$$40x+20-9x^{2}=36$$

Subtract 36 from both sides.

$$40x+20-9x^{2}-36=0$$

Subtract 36 from 20 to get -16.

$$40x-16-9x^{2}=0$$

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

$$-9x^{2}+40x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=40 ab=-9\left(-16\right)=144$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 144.

$$1,144 2,72 3,48 4,36 6,24 8,18 9,16 12,12$$

Calculate the sum for each pair.

$$1+144=145 2+72=74 3+48=51 4+36=40 6+24=30 8+18=26 9+16=25 12+12=24$$

The solution is the pair that gives sum 40.

$$a=36 b=4$$

Rewrite -9x^{2}+40x-16 as \left(-9x^{2}+36x\right)+\left(4x-16\right).

$$\left(-9x^{2}+36x\right)+\left(4x-16\right)$$

Factor out 9x in the first and -4 in the second group.

$$9x\left(-x+4\right)-4\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(9x-4\right)$$

To find equation solutions, solve -x+4=0 and 9x-4=0.

$$x=4 x=\frac{4}{9}$$

Substitute 4 for x in the equation \sqrt{x+5}+1=2\sqrt{x}.

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

Simplify. The value x=4 satisfies the equation.

$$4=4$$

Substitute \frac{4}{9} for x in the equation \sqrt{x+5}+1=2\sqrt{x}.

$$\sqrt{\frac{4}{9}+5}+1=2\sqrt{\frac{4}{9}}$$

Simplify. The value x=\frac{4}{9} does not satisfy the equation.

$$\frac{10}{3}=\frac{4}{3}$$

Substitute 4 for x in the equation \sqrt{x+5}+1=2\sqrt{x}.

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

Simplify. The value x=4 satisfies the equation.

$$4=4$$

Equation \sqrt{x+5}+1=2\sqrt{x} has a unique solution.

$$x=4$$