Solve sqrt{2x-9}=5-sqrt{x-3}

$$\sqrt { 2 x - 9 } = 5 - \sqrt { x - 3 }$$

Solve for x

$$x=81-10\sqrt{53}\approx 8.198901107$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{2x-9} to the power of 2 and get 2x-9.

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

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

$$2x-9=25-10\sqrt{x-3}+\left(\sqrt{x-3}\right)^{2}$$

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

$$2x-9=25-10\sqrt{x-3}+x-3$$

Subtract 3 from 25 to get 22.

$$2x-9=22-10\sqrt{x-3}+x$$

Subtract 22+x from both sides of the equation.

$$2x-9-\left(22+x\right)=-10\sqrt{x-3}$$

To find the opposite of 22+x, find the opposite of each term.

$$2x-9-22-x=-10\sqrt{x-3}$$

Subtract 22 from -9 to get -31.

$$2x-31-x=-10\sqrt{x-3}$$

Combine 2x and -x to get x.

$$x-31=-10\sqrt{x-3}$$

Square both sides of the equation.

$$\left(x-31\right)^{2}=\left(-10\sqrt{x-3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-31\right)^{2}.

$$x^{2}-62x+961=\left(-10\sqrt{x-3}\right)^{2}$$

Expand \left(-10\sqrt{x-3}\right)^{2}.

$$x^{2}-62x+961=\left(-10\right)^{2}\left(\sqrt{x-3}\right)^{2}$$

Calculate -10 to the power of 2 and get 100.

$$x^{2}-62x+961=100\left(\sqrt{x-3}\right)^{2}$$

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

$$x^{2}-62x+961=100\left(x-3\right)$$

Use the distributive property to multiply 100 by x-3.

$$x^{2}-62x+961=100x-300$$

Subtract 100x from both sides.

$$x^{2}-62x+961-100x=-300$$

Combine -62x and -100x to get -162x.

$$x^{2}-162x+961=-300$$

Add 300 to both sides.

$$x^{2}-162x+961+300=0$$

Add 961 and 300 to get 1261.

$$x^{2}-162x+1261=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -162 for b, and 1261 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-162\right)±\sqrt{\left(-162\right)^{2}-4\times 1261}}{2}$$

Square -162.

$$x=\frac{-\left(-162\right)±\sqrt{26244-4\times 1261}}{2}$$

Multiply -4 times 1261.

$$x=\frac{-\left(-162\right)±\sqrt{26244-5044}}{2}$$

Add 26244 to -5044.

$$x=\frac{-\left(-162\right)±\sqrt{21200}}{2}$$

Take the square root of 21200.

$$x=\frac{-\left(-162\right)±20\sqrt{53}}{2}$$

The opposite of -162 is 162.

$$x=\frac{162±20\sqrt{53}}{2}$$

Now solve the equation x=\frac{162±20\sqrt{53}}{2} when ± is plus. Add 162 to 20\sqrt{53}.

$$x=\frac{20\sqrt{53}+162}{2}$$

Divide 162+20\sqrt{53} by 2.

$$x=10\sqrt{53}+81$$

Now solve the equation x=\frac{162±20\sqrt{53}}{2} when ± is minus. Subtract 20\sqrt{53} from 162.

$$x=\frac{162-20\sqrt{53}}{2}$$

Divide 162-20\sqrt{53} by 2.

$$x=81-10\sqrt{53}$$

The equation is now solved.

$$x=10\sqrt{53}+81 x=81-10\sqrt{53}$$

Substitute 10\sqrt{53}+81 for x in the equation \sqrt{2x-9}=5-\sqrt{x-3}.

$$\sqrt{2\left(10\sqrt{53}+81\right)-9}=5-\sqrt{10\sqrt{53}+81-3}$$

Simplify. The value x=10\sqrt{53}+81 does not satisfy the equation because the left and the right hand side have opposite signs.

$$10+53^{\frac{1}{2}}=-53^{\frac{1}{2}}$$

Substitute 81-10\sqrt{53} for x in the equation \sqrt{2x-9}=5-\sqrt{x-3}.

$$\sqrt{2\left(81-10\sqrt{53}\right)-9}=5-\sqrt{81-10\sqrt{53}-3}$$

Simplify. The value x=81-10\sqrt{53} satisfies the equation.

$$10-53^{\frac{1}{2}}=10-53^{\frac{1}{2}}$$

Equation \sqrt{2x-9}=-\sqrt{x-3}+5 has a unique solution.

$$x=81-10\sqrt{53}$$