Solve (8sqrt{x})^2=(4+sqrt{x})^2

$$( 8 \sqrt { x } ) ^ { 2 } = ( 4 + \sqrt { x } ) ^ { 2 }$$

Solve for x

$$x=\frac{16}{49}\approx 0.326530612$$

Solution Steps

Expand \left(8\sqrt{x}\right)^{2}.

$$8^{2}\left(\sqrt{x}\right)^{2}=\left(4+\sqrt{x}\right)^{2}$$

Calculate 8 to the power of 2 and get 64.

$$64\left(\sqrt{x}\right)^{2}=\left(4+\sqrt{x}\right)^{2}$$

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

$$64x=\left(4+\sqrt{x}\right)^{2}$$

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

$$64x=16+8\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$64x=16+8\sqrt{x}+x$$

Subtract 8\sqrt{x} from both sides.

$$64x-8\sqrt{x}=16+x$$

Subtract x from both sides.

$$64x-8\sqrt{x}-x=16$$

Combine 64x and -x to get 63x.

$$63x-8\sqrt{x}=16$$

Subtract 63x from both sides of the equation.

$$-8\sqrt{x}=16-63x$$

Square both sides of the equation.

$$\left(-8\sqrt{x}\right)^{2}=\left(-63x+16\right)^{2}$$

Expand \left(-8\sqrt{x}\right)^{2}.

$$\left(-8\right)^{2}\left(\sqrt{x}\right)^{2}=\left(-63x+16\right)^{2}$$

Calculate -8 to the power of 2 and get 64.

$$64\left(\sqrt{x}\right)^{2}=\left(-63x+16\right)^{2}$$

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

$$64x=\left(-63x+16\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-63x+16\right)^{2}.

$$64x=3969x^{2}-2016x+256$$

Subtract 3969x^{2} from both sides.

$$64x-3969x^{2}=-2016x+256$$

Add 2016x to both sides.

$$64x-3969x^{2}+2016x=256$$

Combine 64x and 2016x to get 2080x.

$$2080x-3969x^{2}=256$$

Subtract 256 from both sides.

$$2080x-3969x^{2}-256=0$$

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.

$$-3969x^{2}+2080x-256=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3969 for a, 2080 for b, and -256 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2080±\sqrt{2080^{2}-4\left(-3969\right)\left(-256\right)}}{2\left(-3969\right)}$$

Square 2080.

$$x=\frac{-2080±\sqrt{4326400-4\left(-3969\right)\left(-256\right)}}{2\left(-3969\right)}$$

Multiply -4 times -3969.

$$x=\frac{-2080±\sqrt{4326400+15876\left(-256\right)}}{2\left(-3969\right)}$$

Multiply 15876 times -256.

$$x=\frac{-2080±\sqrt{4326400-4064256}}{2\left(-3969\right)}$$

Add 4326400 to -4064256.

$$x=\frac{-2080±\sqrt{262144}}{2\left(-3969\right)}$$

Take the square root of 262144.

$$x=\frac{-2080±512}{2\left(-3969\right)}$$

Multiply 2 times -3969.

$$x=\frac{-2080±512}{-7938}$$

Now solve the equation x=\frac{-2080±512}{-7938} when ± is plus. Add -2080 to 512.

$$x=-\frac{1568}{-7938}$$

Reduce the fraction \frac{-1568}{-7938} to lowest terms by extracting and canceling out 98.

$$x=\frac{16}{81}$$

Now solve the equation x=\frac{-2080±512}{-7938} when ± is minus. Subtract 512 from -2080.

$$x=-\frac{2592}{-7938}$$

Reduce the fraction \frac{-2592}{-7938} to lowest terms by extracting and canceling out 162.

$$x=\frac{16}{49}$$

The equation is now solved.

$$x=\frac{16}{81} x=\frac{16}{49}$$

Substitute \frac{16}{81} for x in the equation \left(8\sqrt{x}\right)^{2}=\left(4+\sqrt{x}\right)^{2}.

$$\left(8\sqrt{\frac{16}{81}}\right)^{2}=\left(4+\sqrt{\frac{16}{81}}\right)^{2}$$

Simplify. The value x=\frac{16}{81} does not satisfy the equation.

$$\frac{1024}{81}=\frac{1600}{81}$$

Substitute \frac{16}{49} for x in the equation \left(8\sqrt{x}\right)^{2}=\left(4+\sqrt{x}\right)^{2}.

$$\left(8\sqrt{\frac{16}{49}}\right)^{2}=\left(4+\sqrt{\frac{16}{49}}\right)^{2}$$

Simplify. The value x=\frac{16}{49} satisfies the equation.

$$\frac{1024}{49}=\frac{1024}{49}$$

Equation -8\sqrt{x}=16-63x has a unique solution.

$$x=\frac{16}{49}$$