Solve sqrt{2x-3}+sqrt{4x+1}=4

$$\sqrt{ 2x-3 } + \sqrt{ 4x+1 } =4$$

Solve for x

$$x=2$$

Solution Steps

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

$$\sqrt{2x-3}=4-\sqrt{4x+1}$$

Square both sides of the equation.

$$\left(\sqrt{2x-3}\right)^{2}=\left(4-\sqrt{4x+1}\right)^{2}$$

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

$$2x-3=\left(4-\sqrt{4x+1}\right)^{2}$$

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

$$2x-3=16-8\sqrt{4x+1}+\left(\sqrt{4x+1}\right)^{2}$$

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

$$2x-3=16-8\sqrt{4x+1}+4x+1$$

Add 16 and 1 to get 17.

$$2x-3=17-8\sqrt{4x+1}+4x$$

Subtract 17+4x from both sides of the equation.

$$2x-3-\left(17+4x\right)=-8\sqrt{4x+1}$$

To find the opposite of 17+4x, find the opposite of each term.

$$2x-3-17-4x=-8\sqrt{4x+1}$$

Subtract 17 from -3 to get -20.

$$2x-20-4x=-8\sqrt{4x+1}$$

Combine 2x and -4x to get -2x.

$$-2x-20=-8\sqrt{4x+1}$$

Square both sides of the equation.

$$\left(-2x-20\right)^{2}=\left(-8\sqrt{4x+1}\right)^{2}$$

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

$$4x^{2}+80x+400=\left(-8\sqrt{4x+1}\right)^{2}$$

Expand \left(-8\sqrt{4x+1}\right)^{2}.

$$4x^{2}+80x+400=\left(-8\right)^{2}\left(\sqrt{4x+1}\right)^{2}$$

Calculate -8 to the power of 2 and get 64.

$$4x^{2}+80x+400=64\left(\sqrt{4x+1}\right)^{2}$$

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

$$4x^{2}+80x+400=64\left(4x+1\right)$$

Use the distributive property to multiply 64 by 4x+1.

$$4x^{2}+80x+400=256x+64$$

Subtract 256x from both sides.

$$4x^{2}+80x+400-256x=64$$

Combine 80x and -256x to get -176x.

$$4x^{2}-176x+400=64$$

Subtract 64 from both sides.

$$4x^{2}-176x+400-64=0$$

Subtract 64 from 400 to get 336.

$$4x^{2}-176x+336=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -176 for b, and 336 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-176\right)±\sqrt{\left(-176\right)^{2}-4\times 4\times 336}}{2\times 4}$$

Square -176.

$$x=\frac{-\left(-176\right)±\sqrt{30976-4\times 4\times 336}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-176\right)±\sqrt{30976-16\times 336}}{2\times 4}$$

Multiply -16 times 336.

$$x=\frac{-\left(-176\right)±\sqrt{30976-5376}}{2\times 4}$$

Add 30976 to -5376.

$$x=\frac{-\left(-176\right)±\sqrt{25600}}{2\times 4}$$

Take the square root of 25600.

$$x=\frac{-\left(-176\right)±160}{2\times 4}$$

The opposite of -176 is 176.

$$x=\frac{176±160}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{176±160}{8}$$

Now solve the equation x=\frac{176±160}{8} when ± is plus. Add 176 to 160.

$$x=\frac{336}{8}$$

Divide 336 by 8.

$$x=42$$

Now solve the equation x=\frac{176±160}{8} when ± is minus. Subtract 160 from 176.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

The equation is now solved.

$$x=42 x=2$$

Substitute 42 for x in the equation \sqrt{2x-3}+\sqrt{4x+1}=4.

$$\sqrt{2\times 42-3}+\sqrt{4\times 42+1}=4$$

Simplify. The value x=42 does not satisfy the equation.

$$22=4$$

Substitute 2 for x in the equation \sqrt{2x-3}+\sqrt{4x+1}=4.

$$\sqrt{2\times 2-3}+\sqrt{4\times 2+1}=4$$

Simplify. The value x=2 satisfies the equation.

$$4=4$$

Equation \sqrt{2x-3}=-\sqrt{4x+1}+4 has a unique solution.

$$x=2$$