Solve sqrt{3x^2}-xsqrt{12}+2xsqrt{75}=sqrt{3}

$$\sqrt { 3 x ^ { 2 } } - x \sqrt { 12 } + 2 x \sqrt { 75 } = \sqrt { 3 }$$

Solve for x

$$x=\frac{1}{9}\approx 0.111111111$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\sqrt{3x^{2}}-x\times 2\sqrt{3}+2x\sqrt{75}=\sqrt{3}$$

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

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

Multiply 2 and 5 to get 10.

$$\sqrt{3x^{2}}-x\times 2\sqrt{3}+10x\sqrt{3}=\sqrt{3}$$

Multiply -1 and 2 to get -2.

$$\sqrt{3x^{2}}-2x\sqrt{3}+10x\sqrt{3}=\sqrt{3}$$

Combine -2x\sqrt{3} and 10x\sqrt{3} to get 8x\sqrt{3}.

$$\sqrt{3x^{2}}+8x\sqrt{3}=\sqrt{3}$$

Subtract 8x\sqrt{3} from both sides of the equation.

$$\sqrt{3x^{2}}=\sqrt{3}-8x\sqrt{3}$$

Square both sides of the equation.

$$\left(\sqrt{3x^{2}}\right)^{2}=\left(-8x\sqrt{3}+\sqrt{3}\right)^{2}$$

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

$$3x^{2}=\left(-8x\sqrt{3}+\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-8x\sqrt{3}+\sqrt{3}\right)^{2}.

$$3x^{2}=64x^{2}\left(\sqrt{3}\right)^{2}-16x\sqrt{3}\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3x^{2}=64x^{2}\left(\sqrt{3}\right)^{2}-16x\times 3+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$3x^{2}=64x^{2}\times 3-16x\times 3+\left(\sqrt{3}\right)^{2}$$

Multiply 64 and 3 to get 192.

$$3x^{2}=192x^{2}-16x\times 3+\left(\sqrt{3}\right)^{2}$$

Multiply -16 and 3 to get -48.

$$3x^{2}=192x^{2}-48x+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$3x^{2}=192x^{2}-48x+3$$

Subtract 192x^{2} from both sides.

$$3x^{2}-192x^{2}=-48x+3$$

Combine 3x^{2} and -192x^{2} to get -189x^{2}.

$$-189x^{2}=-48x+3$$

Add 48x to both sides.

$$-189x^{2}+48x=3$$

Subtract 3 from both sides.

$$-189x^{2}+48x-3=0$$

Divide both sides by 3.

$$-63x^{2}+16x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -63x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=16 ab=-63\left(-1\right)=63$$

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 63.

$$1,63 3,21 7,9$$

Calculate the sum for each pair.

$$1+63=64 3+21=24 7+9=16$$

The solution is the pair that gives sum 16.

$$a=9 b=7$$

Rewrite -63x^{2}+16x-1 as \left(-63x^{2}+9x\right)+\left(7x-1\right).

$$\left(-63x^{2}+9x\right)+\left(7x-1\right)$$

Factor out -9x in -63x^{2}+9x.

$$-9x\left(7x-1\right)+7x-1$$

Factor out common term 7x-1 by using distributive property.

$$\left(7x-1\right)\left(-9x+1\right)$$

To find equation solutions, solve 7x-1=0 and -9x+1=0.

$$x=\frac{1}{7} x=\frac{1}{9}$$

Substitute \frac{1}{7} for x in the equation \sqrt{3x^{2}}-x\sqrt{12}+2x\sqrt{75}=\sqrt{3}.

$$\sqrt{3\times \left(\frac{1}{7}\right)^{2}}-\frac{1}{7}\sqrt{12}+2\times \frac{1}{7}\sqrt{75}=\sqrt{3}$$

Simplify. The value x=\frac{1}{7} does not satisfy the equation.

$$\frac{9}{7}\times 3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

Substitute \frac{1}{9} for x in the equation \sqrt{3x^{2}}-x\sqrt{12}+2x\sqrt{75}=\sqrt{3}.

$$\sqrt{3\times \left(\frac{1}{9}\right)^{2}}-\frac{1}{9}\sqrt{12}+2\times \frac{1}{9}\sqrt{75}=\sqrt{3}$$

Simplify. The value x=\frac{1}{9} satisfies the equation.

$$3^{\frac{1}{2}}=3^{\frac{1}{2}}$$

Equation \sqrt{3x^{2}}=-8\sqrt{3}x+\sqrt{3} has a unique solution.

$$x=\frac{1}{9}$$