Solve 1+sqrt{4-3x}=sqrt{9-6x}

$$1 + \sqrt { 4 - 3 x } = \sqrt { 9 - 6 x }$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=0$$

Solution Steps

Square both sides of the equation.

$$\left(1+\sqrt{4-3x}\right)^{2}=\left(\sqrt{9-6x}\right)^{2}$$

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

$$1+2\sqrt{4-3x}+\left(\sqrt{4-3x}\right)^{2}=\left(\sqrt{9-6x}\right)^{2}$$

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

$$1+2\sqrt{4-3x}+4-3x=\left(\sqrt{9-6x}\right)^{2}$$

Add 1 and 4 to get 5.

$$5+2\sqrt{4-3x}-3x=\left(\sqrt{9-6x}\right)^{2}$$

Calculate \sqrt{9-6x} to the power of 2 and get 9-6x.

$$5+2\sqrt{4-3x}-3x=9-6x$$

Subtract 5-3x from both sides of the equation.

$$2\sqrt{4-3x}=9-6x-\left(5-3x\right)$$

To find the opposite of 5-3x, find the opposite of each term.

$$2\sqrt{4-3x}=9-6x-5+3x$$

Subtract 5 from 9 to get 4.

$$2\sqrt{4-3x}=4-6x+3x$$

Combine -6x and 3x to get -3x.

$$2\sqrt{4-3x}=4-3x$$

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Use the distributive property to multiply 4 by 4-3x.

$$16-12x=\left(4-3x\right)^{2}$$

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

$$16-12x=16-24x+9x^{2}$$

Subtract 16 from both sides.

$$16-12x-16=-24x+9x^{2}$$

Subtract 16 from 16 to get 0.

$$-12x=-24x+9x^{2}$$

Add 24x to both sides.

$$-12x+24x=9x^{2}$$

Combine -12x and 24x to get 12x.

$$12x=9x^{2}$$

Subtract 9x^{2} from both sides.

$$12x-9x^{2}=0$$

Factor out x.

$$x\left(12-9x\right)=0$$

To find equation solutions, solve x=0 and 12-9x=0.

$$x=0 x=\frac{4}{3}$$

Substitute 0 for x in the equation 1+\sqrt{4-3x}=\sqrt{9-6x}.

$$1+\sqrt{4-3\times 0}=\sqrt{9-6\times 0}$$

Simplify. The value x=0 satisfies the equation.

$$3=3$$

Substitute \frac{4}{3} for x in the equation 1+\sqrt{4-3x}=\sqrt{9-6x}.

$$1+\sqrt{4-3\times \frac{4}{3}}=\sqrt{9-6\times \frac{4}{3}}$$

Simplify. The value x=\frac{4}{3} satisfies the equation.

$$1=1$$

List all solutions of \sqrt{4-3x}+1=\sqrt{9-6x}.

$$x=0 x=\frac{4}{3}$$