Solve sqrt{x+1}+sqrt{3x+4}=sqrt{5x+9}

$$\sqrt{ x+1 } + \sqrt{ 3x+4 } = \sqrt{ 5x+9 }$$

Solve for x

$$x=0$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+1}+\sqrt{3x+4}\right)^{2}=\left(\sqrt{5x+9}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+1}+\sqrt{3x+4}\right)^{2}.

$$\left(\sqrt{x+1}\right)^{2}+2\sqrt{x+1}\sqrt{3x+4}+\left(\sqrt{3x+4}\right)^{2}=\left(\sqrt{5x+9}\right)^{2}$$

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

$$x+1+2\sqrt{x+1}\sqrt{3x+4}+\left(\sqrt{3x+4}\right)^{2}=\left(\sqrt{5x+9}\right)^{2}$$

Calculate \sqrt{3x+4} to the power of 2 and get 3x+4.

$$x+1+2\sqrt{x+1}\sqrt{3x+4}+3x+4=\left(\sqrt{5x+9}\right)^{2}$$

Combine x and 3x to get 4x.

$$4x+1+2\sqrt{x+1}\sqrt{3x+4}+4=\left(\sqrt{5x+9}\right)^{2}$$

Add 1 and 4 to get 5.

$$4x+5+2\sqrt{x+1}\sqrt{3x+4}=\left(\sqrt{5x+9}\right)^{2}$$

Calculate \sqrt{5x+9} to the power of 2 and get 5x+9.

$$4x+5+2\sqrt{x+1}\sqrt{3x+4}=5x+9$$

Subtract 4x+5 from both sides of the equation.

$$2\sqrt{x+1}\sqrt{3x+4}=5x+9-\left(4x+5\right)$$

To find the opposite of 4x+5, find the opposite of each term.

$$2\sqrt{x+1}\sqrt{3x+4}=5x+9-4x-5$$

Combine 5x and -4x to get x.

$$2\sqrt{x+1}\sqrt{3x+4}=x+9-5$$

Subtract 5 from 9 to get 4.

$$2\sqrt{x+1}\sqrt{3x+4}=x+4$$

Square both sides of the equation.

$$\left(2\sqrt{x+1}\sqrt{3x+4}\right)^{2}=\left(x+4\right)^{2}$$

Expand \left(2\sqrt{x+1}\sqrt{3x+4}\right)^{2}.

$$2^{2}\left(\sqrt{x+1}\right)^{2}\left(\sqrt{3x+4}\right)^{2}=\left(x+4\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x+1}\right)^{2}\left(\sqrt{3x+4}\right)^{2}=\left(x+4\right)^{2}$$

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

$$4\left(x+1\right)\left(\sqrt{3x+4}\right)^{2}=\left(x+4\right)^{2}$$

Calculate \sqrt{3x+4} to the power of 2 and get 3x+4.

$$4\left(x+1\right)\left(3x+4\right)=\left(x+4\right)^{2}$$

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

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

Apply the distributive property by multiplying each term of 4x+4 by each term of 3x+4.

$$12x^{2}+16x+12x+16=\left(x+4\right)^{2}$$

Combine 16x and 12x to get 28x.

$$12x^{2}+28x+16=\left(x+4\right)^{2}$$

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

$$12x^{2}+28x+16=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$12x^{2}+28x+16-x^{2}=8x+16$$

Combine 12x^{2} and -x^{2} to get 11x^{2}.

$$11x^{2}+28x+16=8x+16$$

Subtract 8x from both sides.

$$11x^{2}+28x+16-8x=16$$

Combine 28x and -8x to get 20x.

$$11x^{2}+20x+16=16$$

Subtract 16 from both sides.

$$11x^{2}+20x+16-16=0$$

Subtract 16 from 16 to get 0.

$$11x^{2}+20x=0$$

Factor out x.

$$x\left(11x+20\right)=0$$

To find equation solutions, solve x=0 and 11x+20=0.

$$x=0 x=-\frac{20}{11}$$

Substitute -\frac{20}{11} for x in the equation \sqrt{x+1}+\sqrt{3x+4}=\sqrt{5x+9}. The expression \sqrt{-\frac{20}{11}+1} is undefined because the radicand cannot be negative.

$$\sqrt{-\frac{20}{11}+1}+\sqrt{3\left(-\frac{20}{11}\right)+4}=\sqrt{5\left(-\frac{20}{11}\right)+9}$$

Substitute 0 for x in the equation \sqrt{x+1}+\sqrt{3x+4}=\sqrt{5x+9}.

$$\sqrt{0+1}+\sqrt{3\times 0+4}=\sqrt{5\times 0+9}$$

Simplify. The value x=0 satisfies the equation.

$$3=3$$

Equation \sqrt{x+1}+\sqrt{3x+4}=\sqrt{5x+9} has a unique solution.

$$x=0$$