Solve sqrt{3sqrt{x+1}}=sqrt{3x-5}

$$\sqrt{ 3 \sqrt{ x+1 } } = \sqrt{ 3x-5 }$$

Solve for x

$$x = \frac{\sqrt{105} + 13}{6} \approx 3.874491794$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{3\sqrt{x+1}} to the power of 2 and get 3\sqrt{x+1}.

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

Calculate \sqrt{3x-5} to the power of 2 and get 3x-5.

$$3\sqrt{x+1}=3x-5$$

Square both sides of the equation.

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

Expand \left(3\sqrt{x+1}\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{x+1}\right)^{2}=\left(3x-5\right)^{2}$$

Calculate \sqrt{x+1} to the power of 2 and get x+1.

$$9\left(x+1\right)=\left(3x-5\right)^{2}$$

Use the distributive property to multiply 9 by x+1.

$$9x+9=\left(3x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x+9=9x^{2}-30x+25$$

Subtract 9x^{2} from both sides.

$$9x+9-9x^{2}=-30x+25$$

Add 30x to both sides.

$$9x+9-9x^{2}+30x=25$$

Combine 9x and 30x to get 39x.

$$39x+9-9x^{2}=25$$

Subtract 25 from both sides.

$$39x+9-9x^{2}-25=0$$

Subtract 25 from 9 to get -16.

$$39x-16-9x^{2}=0$$

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.

$$-9x^{2}+39x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 39 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-39±\sqrt{39^{2}-4\left(-9\right)\left(-16\right)}}{2\left(-9\right)}$$

Square 39.

$$x=\frac{-39±\sqrt{1521-4\left(-9\right)\left(-16\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-39±\sqrt{1521+36\left(-16\right)}}{2\left(-9\right)}$$

Multiply 36 times -16.

$$x=\frac{-39±\sqrt{1521-576}}{2\left(-9\right)}$$

Add 1521 to -576.

$$x=\frac{-39±\sqrt{945}}{2\left(-9\right)}$$

Take the square root of 945.

$$x=\frac{-39±3\sqrt{105}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-39±3\sqrt{105}}{-18}$$

Now solve the equation x=\frac{-39±3\sqrt{105}}{-18} when ± is plus. Add -39 to 3\sqrt{105}.

$$x=\frac{3\sqrt{105}-39}{-18}$$

Divide -39+3\sqrt{105} by -18.

$$x=\frac{13-\sqrt{105}}{6}$$

Now solve the equation x=\frac{-39±3\sqrt{105}}{-18} when ± is minus. Subtract 3\sqrt{105} from -39.

$$x=\frac{-3\sqrt{105}-39}{-18}$$

Divide -39-3\sqrt{105} by -18.

$$x=\frac{\sqrt{105}+13}{6}$$

The equation is now solved.

$$x=\frac{13-\sqrt{105}}{6} x=\frac{\sqrt{105}+13}{6}$$

Substitute \frac{13-\sqrt{105}}{6} for x in the equation \sqrt{3\sqrt{x+1}}=\sqrt{3x-5}. The expression \sqrt{3\times \frac{13-\sqrt{105}}{6}-5} is undefined because the radicand cannot be negative.

$$\sqrt{3\sqrt{\frac{13-\sqrt{105}}{6}+1}}=\sqrt{3\times \frac{13-\sqrt{105}}{6}-5}$$

Substitute \frac{\sqrt{105}+13}{6} for x in the equation \sqrt{3\sqrt{x+1}}=\sqrt{3x-5}.

$$\sqrt{3\sqrt{\frac{\sqrt{105}+13}{6}+1}}=\sqrt{3\times \frac{\sqrt{105}+13}{6}-5}$$

Simplify. The value x=\frac{\sqrt{105}+13}{6} satisfies the equation.

$$\left(\frac{3}{2}+\frac{1}{2}\times 105^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(\frac{1}{2}\times 105^{\frac{1}{2}}+\frac{3}{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{3\sqrt{x+1}}=\sqrt{3x-5} has a unique solution.

$$x=\frac{\sqrt{105}+13}{6}$$