Solve x+3=(4x+1)+(x-2)-2sqrt{(4x+1)(x-2)}

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

Solve for x

$$x=6$$

Solution Steps

Subtract 4x+1+x-2 from both sides of the equation.

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

Combine 4x and x to get 5x.

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

Subtract 2 from 1 to get -1.

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

To find the opposite of 5x-1, find the opposite of each term.

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

Combine x and -5x to get -4x.

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

Add 3 and 1 to get 4.

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

Use the distributive property to multiply 4x+1 by x-2 and combine like terms.

$$-4x+4=-2\sqrt{4x^{2}-7x-2}$$

Square both sides of the equation.

$$\left(-4x+4\right)^{2}=\left(-2\sqrt{4x^{2}-7x-2}\right)^{2}$$

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

$$16x^{2}-32x+16=\left(-2\sqrt{4x^{2}-7x-2}\right)^{2}$$

Expand \left(-2\sqrt{4x^{2}-7x-2}\right)^{2}.

$$16x^{2}-32x+16=\left(-2\right)^{2}\left(\sqrt{4x^{2}-7x-2}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$16x^{2}-32x+16=4\left(\sqrt{4x^{2}-7x-2}\right)^{2}$$

Calculate \sqrt{4x^{2}-7x-2} to the power of 2 and get 4x^{2}-7x-2.

$$16x^{2}-32x+16=4\left(4x^{2}-7x-2\right)$$

Use the distributive property to multiply 4 by 4x^{2}-7x-2.

$$16x^{2}-32x+16=16x^{2}-28x-8$$

Subtract 16x^{2} from both sides.

$$16x^{2}-32x+16-16x^{2}=-28x-8$$

Combine 16x^{2} and -16x^{2} to get 0.

$$-32x+16=-28x-8$$

Add 28x to both sides.

$$-32x+16+28x=-8$$

Combine -32x and 28x to get -4x.

$$-4x+16=-8$$

Subtract 16 from both sides.

$$-4x=-8-16$$

Subtract 16 from -8 to get -24.

$$-4x=-24$$

Divide both sides by -4.

$$x=\frac{-24}{-4}$$

Divide -24 by -4 to get 6.

$$x=6$$

Substitute 6 for x in the equation x+3=4x+1+x-2-2\sqrt{\left(4x+1\right)\left(x-2\right)}.

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

Simplify. The value x=6 satisfies the equation.

$$9=9$$

Equation 4-4x=-2\sqrt{4x^{2}-7x-2} has a unique solution.

$$x=6$$