Solve sqrt{6-x}+sqrt{4x+1}=5

$$\sqrt{ 6-x } + \sqrt{ 4x+1 } =5$$

Solve for x

$$x=6$$

$$x=2$$

Solution Steps

Subtract \sqrt{4x+1} from both sides of the equation.

$$\sqrt{6-x}=5-\sqrt{4x+1}$$

Square both sides of the equation.

$$\left(\sqrt{6-x}\right)^{2}=\left(5-\sqrt{4x+1}\right)^{2}$$

Calculate \sqrt{6-x} to the power of 2 and get 6-x.

$$6-x=\left(5-\sqrt{4x+1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{4x+1}\right)^{2}.

$$6-x=25-10\sqrt{4x+1}+\left(\sqrt{4x+1}\right)^{2}$$

Calculate \sqrt{4x+1} to the power of 2 and get 4x+1.

$$6-x=25-10\sqrt{4x+1}+4x+1$$

Add 25 and 1 to get 26.

$$6-x=26-10\sqrt{4x+1}+4x$$

Subtract 26+4x from both sides of the equation.

$$6-x-\left(26+4x\right)=-10\sqrt{4x+1}$$

To find the opposite of 26+4x, find the opposite of each term.

$$6-x-26-4x=-10\sqrt{4x+1}$$

Subtract 26 from 6 to get -20.

$$-20-x-4x=-10\sqrt{4x+1}$$

Combine -x and -4x to get -5x.

$$-20-5x=-10\sqrt{4x+1}$$

Square both sides of the equation.

$$\left(-20-5x\right)^{2}=\left(-10\sqrt{4x+1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-20-5x\right)^{2}.

$$400+200x+25x^{2}=\left(-10\sqrt{4x+1}\right)^{2}$$

Expand \left(-10\sqrt{4x+1}\right)^{2}.

$$400+200x+25x^{2}=\left(-10\right)^{2}\left(\sqrt{4x+1}\right)^{2}$$

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

$$400+200x+25x^{2}=100\left(\sqrt{4x+1}\right)^{2}$$

Calculate \sqrt{4x+1} to the power of 2 and get 4x+1.

$$400+200x+25x^{2}=100\left(4x+1\right)$$

Use the distributive property to multiply 100 by 4x+1.

$$400+200x+25x^{2}=400x+100$$

Subtract 400x from both sides.

$$400+200x+25x^{2}-400x=100$$

Combine 200x and -400x to get -200x.

$$400-200x+25x^{2}=100$$

Subtract 100 from both sides.

$$400-200x+25x^{2}-100=0$$

Subtract 100 from 400 to get 300.

$$300-200x+25x^{2}=0$$

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.

$$25x^{2}-200x+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -200 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-200\right)±\sqrt{\left(-200\right)^{2}-4\times 25\times 300}}{2\times 25}$$

Square -200.

$$x=\frac{-\left(-200\right)±\sqrt{40000-4\times 25\times 300}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-200\right)±\sqrt{40000-100\times 300}}{2\times 25}$$

Multiply -100 times 300.

$$x=\frac{-\left(-200\right)±\sqrt{40000-30000}}{2\times 25}$$

Add 40000 to -30000.

$$x=\frac{-\left(-200\right)±\sqrt{10000}}{2\times 25}$$

Take the square root of 10000.

$$x=\frac{-\left(-200\right)±100}{2\times 25}$$

The opposite of -200 is 200.

$$x=\frac{200±100}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{200±100}{50}$$

Now solve the equation x=\frac{200±100}{50} when ± is plus. Add 200 to 100.

$$x=\frac{300}{50}$$

Divide 300 by 50.

$$x=6$$

Now solve the equation x=\frac{200±100}{50} when ± is minus. Subtract 100 from 200.

$$x=\frac{100}{50}$$

Divide 100 by 50.

$$x=2$$

The equation is now solved.

$$x=6 x=2$$

Substitute 6 for x in the equation \sqrt{6-x}+\sqrt{4x+1}=5.

$$\sqrt{6-6}+\sqrt{4\times 6+1}=5$$

Simplify. The value x=6 satisfies the equation.

$$5=5$$

Substitute 2 for x in the equation \sqrt{6-x}+\sqrt{4x+1}=5.

$$\sqrt{6-2}+\sqrt{4\times 2+1}=5$$

Simplify. The value x=2 satisfies the equation.

$$5=5$$

List all solutions of \sqrt{6-x}=-\sqrt{4x+1}+5.

$$x=6 x=2$$