Solve 2sqrt{9x}-6=10-2sqrt{x}

$$2 \sqrt { 9 x } - 6 = 10 - 2 \sqrt { x }$$

Solve for x

$$x=4$$

Solution Steps

Subtract -6 from both sides of the equation.

$$2\sqrt{9x}=10-2\sqrt{x}+6$$

Square both sides of the equation.

$$\left(2\sqrt{9x}\right)^{2}=\left(10-2\sqrt{x}+6\right)^{2}$$

Expand \left(2\sqrt{9x}\right)^{2}.

$$2^{2}\left(\sqrt{9x}\right)^{2}=\left(10-2\sqrt{x}+6\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{9x}\right)^{2}=\left(10-2\sqrt{x}+6\right)^{2}$$

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

$$4\times 9x=\left(10-2\sqrt{x}+6\right)^{2}$$

Multiply 4 and 9 to get 36.

$$36x=\left(10-2\sqrt{x}+6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10-2\sqrt{x}+6\right)^{2}.

$$36x=\left(10-2\sqrt{x}\right)^{2}+12\left(10-2\sqrt{x}\right)+36$$

Subtract \left(10-2\sqrt{x}\right)^{2} from both sides.

$$36x-\left(10-2\sqrt{x}\right)^{2}=12\left(10-2\sqrt{x}\right)+36$$

Subtract 12\left(10-2\sqrt{x}\right) from both sides.

$$36x-\left(10-2\sqrt{x}\right)^{2}-12\left(10-2\sqrt{x}\right)=36$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-2\sqrt{x}\right)^{2}.

$$36x-\left(100-40\sqrt{x}+4\left(\sqrt{x}\right)^{2}\right)-12\left(10-2\sqrt{x}\right)=36$$

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

$$36x-\left(100-40\sqrt{x}+4x\right)-12\left(10-2\sqrt{x}\right)=36$$

To find the opposite of 100-40\sqrt{x}+4x, find the opposite of each term.

$$36x-100+40\sqrt{x}-4x-12\left(10-2\sqrt{x}\right)=36$$

Combine 36x and -4x to get 32x.

$$32x-100+40\sqrt{x}-12\left(10-2\sqrt{x}\right)=36$$

Use the distributive property to multiply -12 by 10-2\sqrt{x}.

$$32x-100+40\sqrt{x}-120+24\sqrt{x}=36$$

Subtract 120 from -100 to get -220.

$$32x-220+40\sqrt{x}+24\sqrt{x}=36$$

Combine 40\sqrt{x} and 24\sqrt{x} to get 64\sqrt{x}.

$$32x-220+64\sqrt{x}=36$$

Add 220 to both sides.

$$32x+64\sqrt{x}=36+220$$

Add 36 and 220 to get 256.

$$32x+64\sqrt{x}=256$$

Subtract 32x from both sides of the equation.

$$64\sqrt{x}=256-32x$$

Square both sides of the equation.

$$\left(64\sqrt{x}\right)^{2}=\left(-32x+256\right)^{2}$$

Expand \left(64\sqrt{x}\right)^{2}.

$$64^{2}\left(\sqrt{x}\right)^{2}=\left(-32x+256\right)^{2}$$

Calculate 64 to the power of 2 and get 4096.

$$4096\left(\sqrt{x}\right)^{2}=\left(-32x+256\right)^{2}$$

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

$$4096x=\left(-32x+256\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-32x+256\right)^{2}.

$$4096x=1024x^{2}-16384x+65536$$

Subtract 1024x^{2} from both sides.

$$4096x-1024x^{2}=-16384x+65536$$

Add 16384x to both sides.

$$4096x-1024x^{2}+16384x=65536$$

Combine 4096x and 16384x to get 20480x.

$$20480x-1024x^{2}=65536$$

Subtract 65536 from both sides.

$$20480x-1024x^{2}-65536=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-1024x^{2}+20480x-65536=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1024 for a, 20480 for b, and -65536 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20480±\sqrt{20480^{2}-4\left(-1024\right)\left(-65536\right)}}{2\left(-1024\right)}$$

Square 20480.

$$x=\frac{-20480±\sqrt{419430400-4\left(-1024\right)\left(-65536\right)}}{2\left(-1024\right)}$$

Multiply -4 times -1024.

$$x=\frac{-20480±\sqrt{419430400+4096\left(-65536\right)}}{2\left(-1024\right)}$$

Multiply 4096 times -65536.

$$x=\frac{-20480±\sqrt{419430400-268435456}}{2\left(-1024\right)}$$

Add 419430400 to -268435456.

$$x=\frac{-20480±\sqrt{150994944}}{2\left(-1024\right)}$$

Take the square root of 150994944.

$$x=\frac{-20480±12288}{2\left(-1024\right)}$$

Multiply 2 times -1024.

$$x=\frac{-20480±12288}{-2048}$$

Now solve the equation x=\frac{-20480±12288}{-2048} when ± is plus. Add -20480 to 12288.

$$x=-\frac{8192}{-2048}$$

Divide -8192 by -2048.

$$x=4$$

Now solve the equation x=\frac{-20480±12288}{-2048} when ± is minus. Subtract 12288 from -20480.

$$x=-\frac{32768}{-2048}$$

Divide -32768 by -2048.

$$x=16$$

The equation is now solved.

$$x=4 x=16$$

Substitute 4 for x in the equation 2\sqrt{9x}-6=10-2\sqrt{x}.

$$2\sqrt{9\times 4}-6=10-2\sqrt{4}$$

Simplify. The value x=4 satisfies the equation.

$$6=6$$

Substitute 16 for x in the equation 2\sqrt{9x}-6=10-2\sqrt{x}.

$$2\sqrt{9\times 16}-6=10-2\sqrt{16}$$

Simplify. The value x=16 does not satisfy the equation.

$$18=2$$

Substitute 4 for x in the equation 2\sqrt{9x}-6=10-2\sqrt{x}.

$$2\sqrt{9\times 4}-6=10-2\sqrt{4}$$

Simplify. The value x=4 satisfies the equation.

$$6=6$$

Equation 2\sqrt{9x}=10-2\sqrt{x}+6 has a unique solution.

$$x=4$$