Solve 13sqrt{x}=x+36

$$13 \sqrt { x } = x + 36$$

Solve for x

$$x=81$$

$$x=16$$

Solution Steps

Square both sides of the equation.

$$\left(13\sqrt{x}\right)^{2}=\left(x+36\right)^{2}$$

Expand \left(13\sqrt{x}\right)^{2}.

$$13^{2}\left(\sqrt{x}\right)^{2}=\left(x+36\right)^{2}$$

Calculate 13 to the power of 2 and get 169.

$$169\left(\sqrt{x}\right)^{2}=\left(x+36\right)^{2}$$

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

$$169x=\left(x+36\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+36\right)^{2}.

$$169x=x^{2}+72x+1296$$

Subtract x^{2} from both sides.

$$169x-x^{2}=72x+1296$$

Subtract 72x from both sides.

$$169x-x^{2}-72x=1296$$

Combine 169x and -72x to get 97x.

$$97x-x^{2}=1296$$

Subtract 1296 from both sides.

$$97x-x^{2}-1296=0$$

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.

$$-x^{2}+97x-1296=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 97 for b, and -1296 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-97±\sqrt{97^{2}-4\left(-1\right)\left(-1296\right)}}{2\left(-1\right)}$$

Square 97.

$$x=\frac{-97±\sqrt{9409-4\left(-1\right)\left(-1296\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-97±\sqrt{9409+4\left(-1296\right)}}{2\left(-1\right)}$$

Multiply 4 times -1296.

$$x=\frac{-97±\sqrt{9409-5184}}{2\left(-1\right)}$$

Add 9409 to -5184.

$$x=\frac{-97±\sqrt{4225}}{2\left(-1\right)}$$

Take the square root of 4225.

$$x=\frac{-97±65}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-97±65}{-2}$$

Now solve the equation x=\frac{-97±65}{-2} when ± is plus. Add -97 to 65.

$$x=-\frac{32}{-2}$$

Divide -32 by -2.

$$x=16$$

Now solve the equation x=\frac{-97±65}{-2} when ± is minus. Subtract 65 from -97.

$$x=-\frac{162}{-2}$$

Divide -162 by -2.

$$x=81$$

The equation is now solved.

$$x=16 x=81$$

Substitute 16 for x in the equation 13\sqrt{x}=x+36.

$$13\sqrt{16}=16+36$$

Simplify. The value x=16 satisfies the equation.

$$52=52$$

Substitute 81 for x in the equation 13\sqrt{x}=x+36.

$$13\sqrt{81}=81+36$$

Simplify. The value x=81 satisfies the equation.

$$117=117$$

List all solutions of 13\sqrt{x}=x+36.

$$x=16 x=81$$