Solve sqrt{5-x}+sqrt{x}=2.5

$$\sqrt { 5 - x } + \sqrt { x } = 2.5$$

Solve for x

$$x=-\frac{5\sqrt{15}}{8}+2.5\approx 0.079385409$$

$$x=\frac{5\sqrt{15}}{8}+2.5\approx 4.920614591$$

Solution Steps

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

$$\sqrt{5-x}=2.5-\sqrt{x}$$

Square both sides of the equation.

$$\left(\sqrt{5-x}\right)^{2}=\left(2.5-\sqrt{x}\right)^{2}$$

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

$$5-x=\left(2.5-\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2.5-\sqrt{x}\right)^{2}.

$$5-x=6.25-5\sqrt{x}+\left(\sqrt{x}\right)^{2}$$

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

$$5-x=6.25-5\sqrt{x}+x$$

Subtract 6.25+x from both sides of the equation.

$$5-x-\left(6.25+x\right)=-5\sqrt{x}$$

To find the opposite of 6.25+x, find the opposite of each term.

$$5-x-6.25-x=-5\sqrt{x}$$

Subtract 6.25 from 5 to get -1.25.

$$-1.25-x-x=-5\sqrt{x}$$

Combine -x and -x to get -2x.

$$-1.25-2x=-5\sqrt{x}$$

Square both sides of the equation.

$$\left(-1.25-2x\right)^{2}=\left(-5\sqrt{x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-1.25-2x\right)^{2}.

$$1.5625+5x+4x^{2}=\left(-5\sqrt{x}\right)^{2}$$

Expand \left(-5\sqrt{x}\right)^{2}.

$$1.5625+5x+4x^{2}=\left(-5\right)^{2}\left(\sqrt{x}\right)^{2}$$

Calculate -5 to the power of 2 and get 25.

$$1.5625+5x+4x^{2}=25\left(\sqrt{x}\right)^{2}$$

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

$$1.5625+5x+4x^{2}=25x$$

Subtract 25x from both sides.

$$1.5625+5x+4x^{2}-25x=0$$

Combine 5x and -25x to get -20x.

$$1.5625-20x+4x^{2}=0$$

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.

$$4x^{2}-20x+1.5625=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -20 for b, and 1.5625 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 1.5625}}{2\times 4}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 1.5625}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 1.5625}}{2\times 4}$$

Multiply -16 times 1.5625.

$$x=\frac{-\left(-20\right)±\sqrt{400-25}}{2\times 4}$$

Add 400 to -25.

$$x=\frac{-\left(-20\right)±\sqrt{375}}{2\times 4}$$

Take the square root of 375.

$$x=\frac{-\left(-20\right)±5\sqrt{15}}{2\times 4}$$

The opposite of -20 is 20.

$$x=\frac{20±5\sqrt{15}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20±5\sqrt{15}}{8}$$

Now solve the equation x=\frac{20±5\sqrt{15}}{8} when ± is plus. Add 20 to 5\sqrt{15}.

$$x=\frac{5\sqrt{15}+20}{8}$$

Divide 20+5\sqrt{15} by 8.

$$x=\frac{5\sqrt{15}}{8}+\frac{5}{2}$$

Now solve the equation x=\frac{20±5\sqrt{15}}{8} when ± is minus. Subtract 5\sqrt{15} from 20.

$$x=\frac{20-5\sqrt{15}}{8}$$

Divide 20-5\sqrt{15} by 8.

$$x=-\frac{5\sqrt{15}}{8}+\frac{5}{2}$$

The equation is now solved.

$$x=\frac{5\sqrt{15}}{8}+\frac{5}{2} x=-\frac{5\sqrt{15}}{8}+\frac{5}{2}$$

Substitute \frac{5\sqrt{15}}{8}+\frac{5}{2} for x in the equation \sqrt{5-x}+\sqrt{x}=2.5.

$$\sqrt{5-\left(\frac{5\sqrt{15}}{8}+\frac{5}{2}\right)}+\sqrt{\frac{5\sqrt{15}}{8}+\frac{5}{2}}=2.5$$

Simplify. The value x=\frac{5\sqrt{15}}{8}+\frac{5}{2} satisfies the equation.

$$\frac{5}{2}=2.5$$

Substitute -\frac{5\sqrt{15}}{8}+\frac{5}{2} for x in the equation \sqrt{5-x}+\sqrt{x}=2.5.

$$\sqrt{5-\left(-\frac{5\sqrt{15}}{8}+\frac{5}{2}\right)}+\sqrt{-\frac{5\sqrt{15}}{8}+\frac{5}{2}}=2.5$$

Simplify. The value x=-\frac{5\sqrt{15}}{8}+\frac{5}{2} satisfies the equation.

$$\frac{5}{2}=2.5$$

List all solutions of \sqrt{5-x}=-\sqrt{x}+2.5.

$$x=\frac{5\sqrt{15}}{8}+\frac{5}{2} x=-\frac{5\sqrt{15}}{8}+\frac{5}{2}$$