Solve sqrt{3x}+52=x+4

$$\sqrt { 3 x } + 52 = x + 4$$

Solve for x

$$x = \frac{3 {(\sqrt{65} + 33)}}{2} \approx 61.593386622$$

Solution Steps

Subtract 52 from both sides of the equation.

$$\sqrt{3x}=x+4-52$$

Subtract 52 from 4 to get -48.

$$\sqrt{3x}=x-48$$

Square both sides of the equation.

$$\left(\sqrt{3x}\right)^{2}=\left(x-48\right)^{2}$$

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

$$3x=\left(x-48\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-48\right)^{2}.

$$3x=x^{2}-96x+2304$$

Subtract x^{2} from both sides.

$$3x-x^{2}=-96x+2304$$

Add 96x to both sides.

$$3x-x^{2}+96x=2304$$

Combine 3x and 96x to get 99x.

$$99x-x^{2}=2304$$

Subtract 2304 from both sides.

$$99x-x^{2}-2304=0$$

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.

$$-x^{2}+99x-2304=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 99 for b, and -2304 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-99±\sqrt{99^{2}-4\left(-1\right)\left(-2304\right)}}{2\left(-1\right)}$$

Square 99.

$$x=\frac{-99±\sqrt{9801-4\left(-1\right)\left(-2304\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-99±\sqrt{9801+4\left(-2304\right)}}{2\left(-1\right)}$$

Multiply 4 times -2304.

$$x=\frac{-99±\sqrt{9801-9216}}{2\left(-1\right)}$$

Add 9801 to -9216.

$$x=\frac{-99±\sqrt{585}}{2\left(-1\right)}$$

Take the square root of 585.

$$x=\frac{-99±3\sqrt{65}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-99±3\sqrt{65}}{-2}$$

Now solve the equation x=\frac{-99±3\sqrt{65}}{-2} when ± is plus. Add -99 to 3\sqrt{65}.

$$x=\frac{3\sqrt{65}-99}{-2}$$

Divide -99+3\sqrt{65} by -2.

$$x=\frac{99-3\sqrt{65}}{2}$$

Now solve the equation x=\frac{-99±3\sqrt{65}}{-2} when ± is minus. Subtract 3\sqrt{65} from -99.

$$x=\frac{-3\sqrt{65}-99}{-2}$$

Divide -99-3\sqrt{65} by -2.

$$x=\frac{3\sqrt{65}+99}{2}$$

The equation is now solved.

$$x=\frac{99-3\sqrt{65}}{2} x=\frac{3\sqrt{65}+99}{2}$$

Substitute \frac{99-3\sqrt{65}}{2} for x in the equation \sqrt{3x}+52=x+4.

$$\sqrt{3\times \frac{99-3\sqrt{65}}{2}}+52=\frac{99-3\sqrt{65}}{2}+4$$

Simplify. The value x=\frac{99-3\sqrt{65}}{2} does not satisfy the equation.

$$\frac{101}{2}+\frac{3}{2}\times 65^{\frac{1}{2}}=\frac{107}{2}-\frac{3}{2}\times 65^{\frac{1}{2}}$$

Substitute \frac{3\sqrt{65}+99}{2} for x in the equation \sqrt{3x}+52=x+4.

$$\sqrt{3\times \frac{3\sqrt{65}+99}{2}}+52=\frac{3\sqrt{65}+99}{2}+4$$

Simplify. The value x=\frac{3\sqrt{65}+99}{2} satisfies the equation.

$$\frac{107}{2}+\frac{3}{2}\times 65^{\frac{1}{2}}=\frac{3}{2}\times 65^{\frac{1}{2}}+\frac{107}{2}$$

Equation \sqrt{3x}=x-48 has a unique solution.

$$x=\frac{3\sqrt{65}+99}{2}$$