Solve (4x-4)sqrt{2x}=6x-1

$$( 4 x - 4 ) \sqrt { 2 x } = 6 x - 1$$

Solve for x

$$x = \frac{5 \sqrt{17} + 21}{16} \approx 2.600970508$$

$$x=\frac{21-5\sqrt{17}}{16}\approx 0.024029492$$

Solution Steps

Square both sides of the equation.

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

Use the distributive property to multiply 4x-4 by \sqrt{2x}.

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

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

$$16x^{2}\left(\sqrt{2x}\right)^{2}-32x\sqrt{2x}\sqrt{2x}+16\left(\sqrt{2x}\right)^{2}=\left(6x-1\right)^{2}$$

Multiply \sqrt{2x} and \sqrt{2x} to get \left(\sqrt{2x}\right)^{2}.

$$16x^{2}\left(\sqrt{2x}\right)^{2}-32x\left(\sqrt{2x}\right)^{2}+16\left(\sqrt{2x}\right)^{2}=\left(6x-1\right)^{2}$$

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

$$16x^{2}\times 2x-32x\left(\sqrt{2x}\right)^{2}+16\left(\sqrt{2x}\right)^{2}=\left(6x-1\right)^{2}$$

Multiply 16 and 2 to get 32.

$$32x^{2}x-32x\left(\sqrt{2x}\right)^{2}+16\left(\sqrt{2x}\right)^{2}=\left(6x-1\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$32x^{3}-32x\left(\sqrt{2x}\right)^{2}+16\left(\sqrt{2x}\right)^{2}=\left(6x-1\right)^{2}$$

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

$$32x^{3}-32x\times 2x+16\left(\sqrt{2x}\right)^{2}=\left(6x-1\right)^{2}$$

Multiply -32 and 2 to get -64.

$$32x^{3}-64xx+16\left(\sqrt{2x}\right)^{2}=\left(6x-1\right)^{2}$$

Multiply x and x to get x^{2}.

$$32x^{3}-64x^{2}+16\left(\sqrt{2x}\right)^{2}=\left(6x-1\right)^{2}$$

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

$$32x^{3}-64x^{2}+16\times 2x=\left(6x-1\right)^{2}$$

Multiply 16 and 2 to get 32.

$$32x^{3}-64x^{2}+32x=\left(6x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-1\right)^{2}.

$$32x^{3}-64x^{2}+32x=36x^{2}-12x+1$$

Subtract 36x^{2} from both sides.

$$32x^{3}-64x^{2}+32x-36x^{2}=-12x+1$$

Combine -64x^{2} and -36x^{2} to get -100x^{2}.

$$32x^{3}-100x^{2}+32x=-12x+1$$

Add 12x to both sides.

$$32x^{3}-100x^{2}+32x+12x=1$$

Combine 32x and 12x to get 44x.

$$32x^{3}-100x^{2}+44x=1$$

Subtract 1 from both sides.

$$32x^{3}-100x^{2}+44x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 32. List all candidates \frac{p}{q}.

$$±\frac{1}{32},±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 32x^{3}-100x^{2}+44x-1 by 2\left(x-\frac{1}{2}\right)=2x-1 to get 16x^{2}-42x+1. Solve the equation where the result equals to 0.

$$16x^{2}-42x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 16 for a, -42 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}-4\times 16\times 1}}{2\times 16}$$

Do the calculations.

$$x=\frac{42±10\sqrt{17}}{32}$$

Solve the equation 16x^{2}-42x+1=0 when ± is plus and when ± is minus.

$$x=\frac{21-5\sqrt{17}}{16} x=\frac{5\sqrt{17}+21}{16}$$

List all found solutions.

$$x=\frac{1}{2} x=\frac{21-5\sqrt{17}}{16} x=\frac{5\sqrt{17}+21}{16}$$

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

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

Simplify. The value x=\frac{1}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Substitute \frac{21-5\sqrt{17}}{16} for x in the equation \left(4x-4\right)\sqrt{2x}=6x-1.

$$\left(4\times \frac{21-5\sqrt{17}}{16}-4\right)\sqrt{2\times \frac{21-5\sqrt{17}}{16}}=6\times \frac{21-5\sqrt{17}}{16}-1$$

Simplify. The value x=\frac{21-5\sqrt{17}}{16} satisfies the equation.

$$\frac{55}{8}-\frac{15}{8}\times 17^{\frac{1}{2}}=\frac{55}{8}-\frac{15}{8}\times 17^{\frac{1}{2}}$$

Substitute \frac{5\sqrt{17}+21}{16} for x in the equation \left(4x-4\right)\sqrt{2x}=6x-1.

$$\left(4\times \frac{5\sqrt{17}+21}{16}-4\right)\sqrt{2\times \frac{5\sqrt{17}+21}{16}}=6\times \frac{5\sqrt{17}+21}{16}-1$$

Simplify. The value x=\frac{5\sqrt{17}+21}{16} satisfies the equation.

$$\frac{55}{8}+\frac{15}{8}\times 17^{\frac{1}{2}}=\frac{15}{8}\times 17^{\frac{1}{2}}+\frac{55}{8}$$

List all solutions of \sqrt{2x}\left(4x-4\right)=6x-1.

$$x=\frac{21-5\sqrt{17}}{16} x=\frac{5\sqrt{17}+21}{16}$$