Solve sqrt{2x}+sqrt{x-1}=5

$$\sqrt { 2 x } + \sqrt { x - 1 } = 5$$

Solve for x

$$x=74-40\sqrt{3}\approx 4.717967697$$

Solution Steps

Subtract \sqrt{x-1} from both sides of the equation.

$$\sqrt{2x}=5-\sqrt{x-1}$$

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{x-1}\right)^{2}.

$$2x=25-10\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}$$

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

$$2x=25-10\sqrt{x-1}+x-1$$

Subtract 1 from 25 to get 24.

$$2x=24-10\sqrt{x-1}+x$$

Subtract 24+x from both sides of the equation.

$$2x-\left(24+x\right)=-10\sqrt{x-1}$$

To find the opposite of 24+x, find the opposite of each term.

$$2x-24-x=-10\sqrt{x-1}$$

Combine 2x and -x to get x.

$$x-24=-10\sqrt{x-1}$$

Square both sides of the equation.

$$\left(x-24\right)^{2}=\left(-10\sqrt{x-1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-24\right)^{2}.

$$x^{2}-48x+576=\left(-10\sqrt{x-1}\right)^{2}$$

Expand \left(-10\sqrt{x-1}\right)^{2}.

$$x^{2}-48x+576=\left(-10\right)^{2}\left(\sqrt{x-1}\right)^{2}$$

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

$$x^{2}-48x+576=100\left(\sqrt{x-1}\right)^{2}$$

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

$$x^{2}-48x+576=100\left(x-1\right)$$

Use the distributive property to multiply 100 by x-1.

$$x^{2}-48x+576=100x-100$$

Subtract 100x from both sides.

$$x^{2}-48x+576-100x=-100$$

Combine -48x and -100x to get -148x.

$$x^{2}-148x+576=-100$$

Add 100 to both sides.

$$x^{2}-148x+576+100=0$$

Add 576 and 100 to get 676.

$$x^{2}-148x+676=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -148 for b, and 676 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-148\right)±\sqrt{\left(-148\right)^{2}-4\times 676}}{2}$$

Square -148.

$$x=\frac{-\left(-148\right)±\sqrt{21904-4\times 676}}{2}$$

Multiply -4 times 676.

$$x=\frac{-\left(-148\right)±\sqrt{21904-2704}}{2}$$

Add 21904 to -2704.

$$x=\frac{-\left(-148\right)±\sqrt{19200}}{2}$$

Take the square root of 19200.

$$x=\frac{-\left(-148\right)±80\sqrt{3}}{2}$$

The opposite of -148 is 148.

$$x=\frac{148±80\sqrt{3}}{2}$$

Now solve the equation x=\frac{148±80\sqrt{3}}{2} when ± is plus. Add 148 to 80\sqrt{3}.

$$x=\frac{80\sqrt{3}+148}{2}$$

Divide 148+80\sqrt{3} by 2.

$$x=40\sqrt{3}+74$$

Now solve the equation x=\frac{148±80\sqrt{3}}{2} when ± is minus. Subtract 80\sqrt{3} from 148.

$$x=\frac{148-80\sqrt{3}}{2}$$

Divide 148-80\sqrt{3} by 2.

$$x=74-40\sqrt{3}$$

The equation is now solved.

$$x=40\sqrt{3}+74 x=74-40\sqrt{3}$$

Substitute 40\sqrt{3}+74 for x in the equation \sqrt{2x}+\sqrt{x-1}=5.

$$\sqrt{2\left(40\sqrt{3}+74\right)}+\sqrt{40\sqrt{3}+74-1}=5$$

Simplify. The value x=40\sqrt{3}+74 does not satisfy the equation.

$$15+8\times 3^{\frac{1}{2}}=5$$

Substitute 74-40\sqrt{3} for x in the equation \sqrt{2x}+\sqrt{x-1}=5.

$$\sqrt{2\left(74-40\sqrt{3}\right)}+\sqrt{74-40\sqrt{3}-1}=5$$

Simplify. The value x=74-40\sqrt{3} satisfies the equation.

$$5=5$$

Equation \sqrt{2x}=-\sqrt{x-1}+5 has a unique solution.

$$x=74-40\sqrt{3}$$